QED Monthly Volume 2qedmonthly.weebly.com/uploads/2/5/5/2/25525553/qed...QED Monthly Volume 2...

QED Monthly Volume 2 Alexander Katz

QED Monthly Volume 2

Authored by Alexander Katz

Contents

1 AIME Algebra Parts I and II 3

2 AIME Combinatorics Part I 25

3 AIME Combinatorics Part II 36

4 AIME Geometry Part I 46

5 AIME Geometry Part II 55

6 AIME Number Theory Part I 67

7 AIME Number Theory Part II 77

8 NEW! Contests and monthly leaderboards 87

9 Credits Mock AIME Information 89

2

1

AIME Algebra Parts I and II

In a great deal of (primarily Algebra) problems, we make use of polynomi-als. We’ll breeze through the definitions relatively quickly, as it is practicallyimpossible to not have experienced polynomials in some form previously.

Definition 1.0.1. A polynomial is an expression of the form

P (x) =n∑i=0

aixi = anx

n + an−1xn−1 + . . .+ a0,

where x is a variable and n, also known as the degree of the polynomial,is some finite nonnegative integer. We sometimes refer to the degree of apolynomial P as deg(P ). Each of the ai are constant coefficients, usuallyreal numbers. an is often referred to as the leading coefficient, and thepolynomial is said to be monic if an = 1. Moreover, the solutions to theequation P (x) = 0 are called the roots (or sometimes zeros or solutionsof the polynomial. Notice that these roots are often nonreal.

For example, x + 3, x2 + 17, π27x3 + e2x2 + 1337x + 2, and 5 are allexamples of polynomials, while 1

x, 1−x2

x3, and x2−1

x−2 are not.

P (x) Degree, or deg(P ) Leading coefficient Rootsx+3 1 1 (monic) -3

x2 + 17 2 1 (monic) i√

17 and −i√

17π27x3 + e2x2 + 1337x+ 2 3 π27 Who knows?

x2 + 2x+ 1 2 1 (monic) -1 and -1 (!)3 0 3 None

3


When we talk about polynomials, it is important to remember that weare essentially dealing with elementary examples of functions. For example,if P (x) = x2 − 1, then P (2) = 22 − 1 = 3 and P (100) = 1002 − 1 = 9999.We can plug in any convenient value for x we choose (within the domain ofthe polynomial of course, typically R) and return another value.

Polynomials and integers share several similar properties. Just like wecan add two numbers, we can easily add two polynomials. The key point isrealizing that a1x

n+a2xn = (a1+a2)x

n, such as 2x2+3x2 = 5x2. Subtractingpolynomials, as usual, is exactly the same process as addition. Multiplyingpolynomials is only slightly more complicated - we need to make use ofthe distributive property and the fact that a1x

n · a2xm = (a1a2)xm+n. For

example,

(x2+1)(x+1) = x2(x+1)+1(x+1) = x2 ·x+x2 ·1+1·x+1·1 = x3+x2+x+1.

But most importantly, we can divide polynomials in a very similar wayto how we divide integers. Sometimes this is easy - just like the quotient of20 and 5 is 4, the quotient of x2−1 and x−1 is simply x+1. However, not allpolynomials are multiples of one another. Just as 19

5is not an integer, x2+2

x−1is not a polynomial. Despite this, in the same way we can write 19 = 5·3+4,we can write x2 + 2 = (x + 1)(x − 1) + 3. For integers, we can write thismore generally as

Theorem 1.0.1. Given integers a and b, there exist unique integers cand d such that a = bc+ d and 0 ≤ d < b.

Proving this is hardly a difficult exercise - this is simply the division processat work. Notice that we stipulate 0 ≤ d < b to ensure the unicity of thesolution, otherwise we could write 19 = 5 · 1 + 14 = 5 · 2 + 9 = . . . and getinfinite solutions. Unsurprisingly, polynomials work in almost exactly thesame way:

Theorem 1.0.2. Given polynomials f(x) and g(x), there exist uniquepolynomials q(x) and r(x) such that f(x) = g(x)q(x) + r(x) and 0 ≤deg(r) < deg(g)

When we talk about integers, we often want to refer to their prime factor-ization. For example, 15 = 3 · 5, 21 = 3 · 7, 2014 = 2 · 19 · 53, and so on.


Similarly, we often want to know the factorization of polynomials. An obvi-ous problem arises - what is the polynomial equivalent of a prime number?For example, we can write x3 + x2 + x + 1 = (x2 + 1)(x + 1) (we saw thisbefore), but also as (x − i)(x + i)(x + 1). Which one is more valid? Theshort answer is neither. Each have their own advantages, but practically itis usually most convenient to deal with only those polynomials that havereal coefficients. In this way we can define the equivalent of a prime:

Definition 1.0.2. A polynomial P (x) is irreducible (to be technical, overR) if there do not exist non-constant polynomials Q(x) and R(x), bothwith real coefficients, such that P (x) = Q(x)R(x).

P (x) Irreducible? Factorizationx+ 3 Yes x+ 3x2 + 17 Yes x2 + 17

x3 + x2 + x+ 1 No (x2 + 1)(x+ 1)x2 + 2x+ 1 No (x+ 1)2

2x2 + 2 Yes 2x2 + 2 or 2(x2 + 1)

Notice that we essentially ignore constants, because the polynomial 2P (x)behaves in extremely similar ways to P (x). We’ll see more about this later.

Finally, in the same way that prime factorizations of numbers are uniqueup to rearrangement by the Fundamental Theorem of Arithmetic, we cansay the following:

Theorem 1.0.3. For any polynomial P (x), there exists a constant aand unique (up to order) irreducible polynomials Q1(x), Q2(x), . . . , Qk(x)such that

P (x) = aQ1(x)Q2(x) . . . Qk(x).

All of this is great, but just like with integers none of this gives usany information as to how to actually factor a polynomial. Just like we canquickly compute 2·17·37·101 = 127058 (using a calculator if necessary), butfactoring the number 127058 without knowing this information beforehandis practically infeasible, we need to find better ways to tackle polynomialsthan blind guessing. Let’s take another look at the division algorithm forpolynomials:

f(x) = g(x)q(x) + r(x)

What happens when g is a linear polynomial, say x−a. (in other words, weare dividing a polynomial f(x) by a linear polynomial)? Because we have


0 ≤ deg(r) < deg(g) = 1, we must have deg(r) = 0, and so r is a constantpolynomial. Thus we can write r(x) = c, and so f(x) = g(x)(x−a)+ c. Butif we plug in x = a, we get

f(a) = g(a)(a− a) + c = g(a)(0) + c = c

and so we can write f(x) = g(x)(x − a) + f(a) for all a. This gives us thefollowing theorem:

Theorem 1.0.4 (Remainder Theorem). The remainder upon dividingP (x) by x− a is precisely P (a).

This also doesn’t seem too interesting, but this indirectly gives us animportant relation between the roots of a polynomial and its factors. Iff(a) = 0, this implies that f(x) = g(x)(x− a), or x− a is a factor of f(x)!Conversely, if x− a is a factor of f(x), then this implies f(a) = 0. In otherwords,

Theorem 1.0.5 (Factor Theorem). If r is a root of P (x), then x − rdivides P (x). If P (x) is divisible by x− r, then r is a root of P (x).

Notice that when P (x) is divisible by x − r ,this means we can writeP (x) = (x − r)Q(x), where Q is a polynomial (with real coefficients) thathas degree one less than P . If we allow roots to be complex, every non-constant polynomial has a root (this is the Fundamental Theorem ofAlgebra, but intuitively obvious). In fact,

Theorem 1.0.6 (Identity Theorem). The polynomial P (x) has exactlydeg(P ) complex roots. The exception is P (x) = 0, which has infiniteroots. We often refer to the degree of the zero polynomial as −∞, sothat we don’t have to continously make an exception for it. As such, ifa polynomial P (x) has more roots than its degree, the polynomial mustbe identically 0.

In other words, if P (x) is a polynomial of degree n with roots r1, r2, . . . , rn,then we can write

P (x) = an(x− r1)(x− r2) . . . (x− rn)


where a is the leading coefficient, by repeated applications of the factortheorem.

Enough of all this theoretical mumbo jumbo, let’s see how all of thistheory applies to actual problems!

Problem 1.0.1. For what real values of k do 1988x2 + kx + 8891 and8891x2 + kx+ 1988 have a common zero? (Canada, 1988)

Solution. Let this common root be r. The condition tells us that 1988r2 +ra + 8891 = 0 and 8891r2 + ra + 1988 = 0. If we subtract these equations,we get (8891 − 1988)r2 − (8891 − 1988) = 0 =⇒ r2 = 1 =⇒ r = ±1.Plugging this back in gives either

1988(1)2 + k(1) + 8891 = 0 =⇒ x = 10879

or1988(−1)2 + k(−1) + 8891 = 0 =⇒ x = −10879

Problem 1.0.2. A polynomial has remainder 2 when divided by x− 1, andremainder 1 when divided by x− 2. What remainder is obtained when thispolynomial is divided by (x−1)(x−2)? (Lehigh 2011, among other sources)

Solution. Let’s call the polynomial P (x). Notice that the conditions aretelling us P (1) = 2 and P (2) = 1 (if you don’t see this, write out P (x) =Q(x)(x− 1) + 2 and plug in x = 1, doing the same for the other condition).Now we want to find the correct R(x) such that

P (x) = (x− 1)(x− 2)Q(x) +R(x)

Because we are dividing by (x − 1)(x − 2), the degree of R is less than 2.As such, R is either linear or constant, so we can write R = ax+ b for somea, b (notice that we could have a = 0, in which case R would be constant).Plugging in x = 1 and x = 2 gives R(1) = P (1) = 2 and R(2) = P (2) = 1respectively. Thus we have

a+ b = 2, 2a+ b = 1 =⇒ a = −1, b = 3 =⇒ R(x) = 3− x

which is our answer.

Problem 1.0.3. Suppose P (x) = 4(x− 2)2013 + 3. Compute the sum of thecoefficients of P (x).


Solution. At first glance, this seems like an impossible task, or at least onethat is not particuarly Algebra related. We might, for example, wish tomake use of a contrived, practically useless binomial identity. But we havea significiantly better way to do this. Recall the definition of a polynomial:

P (x) = anxn + an−1x

n−1 + . . .+ a0

If we want to find an + an−1 + . . . + a0, the method is now obvious. Wesimply take

P (1) = an1n + an−11n−1 . . .+ a0 = an + an−1 + . . .+ a0.

Finishing the problem, the sum of the coefficients is P (1) = 4(1−2)2013+3 =4(−1)2013 + 3 = 4(−1) + 3 = −1.

However, this gives rise to a natural question - is there a general way todetermine the coefficients of a polynomial? The answer, unsurprisingly, isyes. We have already seen the importance of the roots of polynomials, butif we examine the factored form of a polynomial

P (x) = a(x− r1)(x− r2) . . . (x− rn)

we immediately see some useful connections. For example, the constant termis a(−1)nr1r2 . . . rn, since it is the product of a,−r1,−r2, . . . ,−rn. We canexpand this line of thought into Vieta’s formulas:

Theorem 1.0.7 (Vieta’s Formulas). Define the kth symmetric sum,σk, of a set to be the sum of the elements multiplied k at a time. Forexample, the symmetric sums of {r1, r2, r3} are σ1 = r1 + r2 + r3, σ2 =r1r2 + r1r3 + r2r3, σ3 = r1r2r3. Then

an−k = (−1)k · σk · an

for all 1 ≤ k ≤ n.

Informally, this isn’t too difficult to see. To get a term of the form something·xn−k, we need to choose x from a(x− r1) . . . (x− rn) n− k times, meaningthat we choose −ri k times. But we can, as usual, prove this formally byinduction.


Proof. We will proceed by induction on n. When n = 1, we write P (x) =a1x + a0. The only root of this is −a0

a1, which is just as expected. Assume

that the result is true for n = m. When n = m+ 1 we have

P (x) = am+1xm+1 + amx

m + . . .+ a0

= am+1(x− rm+1)g(x)

where g(x) is a polynomial with roots r1, r2, . . . , rm. By the inductive hy-pothesis, we can write

g(x) = xm − σ1xm−1 + . . .+ (−1)nr1r2 . . . rm

=⇒ f(x) = am+1(x− rm+1)xm − σ1xm−1 + . . .+ (−1)mr1r2 . . . rm.

Multiplying this out gives

f(x) = am+1(xm+1−(r1+r2+ . . .+rm+rm+1)x

m+ . . .+(−1)m+1r1r2 . . . rm+1

which is exactly as desired (each term is σk + rm+1σk−1, which is preciselyσk of the set {r1, r2, . . . , rm+1} as we desire).

The importance of this result cannot be overstated. Many problems re-lated to polynomials are in fact (often not-so-)cleverly disguised applicationsof Vieta’s formulas, often combined with other results.

Problem 1.0.4. Find the sum of the roots, real and non-real, of the equa-

tion x2001 +(12− x)2001

= 0, given that there are no multiple roots. (AIME2001)

Solution. To find the sum of the roots, we only need to find the first 2coefficients of the resulting polynomial. Notice that

x2001 +

(1

2− x)2001

= x2001 − x2001 +1

2

(2001

1

)x2000 − 1

4

(2001

2

)x1999 + stuff

=1

2

(2001

1

)x2000 − 1

4

(2001

2

)x1999 + stuff

Now this is a direct application of our formula. The sum of the roots is

14

(20012

)12

(20011

) =14· 2001·2000

212· 2001

= 500


We could also take advantage of the fact that if x s a root of this poly-nomial, so is 1

2− x. As such there are 1000 pairs of solutions, each of which

sum to 12, and so their total sum is 500. Vieta’s formulas, however, provide

a more general method (though this concept of transforming polynomials isextremely important and will be seen several more times in various prob-lems).

Oftentimes Vieta’s are not sufficient by themselves - we require (some-times significant) other observations, used in tandem, to complete a problem.We’ll go through a couple of examples, starting with a complicated one:

Problem 1.0.5. A polynomial P (x) has real roots can can be written asP (x) = x10−10x9+45x8+a7x

7+a6x6+a5x

5+a4x4+a3x

3+a2x2+a1x+a0, for

some constants a0, a1, . . . , a7. Compute a0 + a1 + a2 + . . .+ a7. (Adaptationof AMSP problem, likely an amalgation of 2003 APMO)

Solution. At first glance, this hardly seems possible. We’re given just 3coefficients of some polynomial, and expected to find the remaining 8? Un-surprisingly, we need to use the roots. The polynomial is of degree 10, so ithas exactly 10 roots (not necessarily distinct). Let’s call those r1, r2, . . . , r10.Then Vieta’s formula’s tell us

r1 + r2 + . . .+ r10 = 10

r1r2 + r1r3 + . . .+ r1r10 + r2r3 + . . .+ r9r10 =∑

1≤i<j≤10

rirj = 45

But now we need to figure out how to continue. We should recognize, atthis point, the standard trick of utilizing the identity

(a1 + a2 + . . .+ an)2 = a21 + a22 + . . .+ a2n + 2∑

1≤i<j≤n

aiaj.

Applying this to our situation, we have

(r1+r2+. . .+rn)2 = 100 = r21+r22+. . .+r2n+2∑

1≤i<j≤10

= r21+r22+. . .+r2n+90

=⇒ r21 + r22 + . . .+ r2n = 10

Notice that we also have r1 + r2 + . . .+ rn = 10, so we can write

(r21 + r22 + . . .+ r210)− 2(r1 + r2 + . . .+ r10) + (1 + 1 + . . .+ 1) = 0


=⇒ (r1 − 1)2 + (r2 − 1)2 + . . .+ (r10 − 1)2 = 0

=⇒ r1 = r2 = . . . = r10 = 1

Therefore, our polynomial is P (x) = (x − 1)10. Recall from a previousexample that the sum of the coefficients is P (1), so we have

a0 + a1 + . . .+ a7 + 45− 10 + 1 = 0 =⇒ a0 + a1 + . . .+ a7 = −36

Let’s look at an easier example, as it provides an excellent segue intoanother important topic.

Problem 1.0.6. Let a, b, and c be the roots of x3 − 3x+ 1 = 0. Find

1

2− a+

1

2− b+

1

2− c.

Solution. Nothing clever about this solution - let’s just put everything overa common denominator and expand:

1

2− a+

1

2− b+

1

2− c

=(2− b)(2− c) + (2− a)(2− c) + (2− a)(2− b)

(2− a)(2− b)(2− c)

=bc− 2b− 2c+ 4 + ac− 2a− 2c+ 4 + ab− 2a− 2b+ 4

−abc+ 2ab+ 2bc+ 2ac− 4a− 4b− 4c+ 8

Now we use Vieta’s formulas, which tell us a + b + c = 0, ab + bc + ac =−3, abc = −1, giving

ab+ bc+ ac− 4(a+ b+ c) + 12

−abc+ 2(ab+ bc+ ac)− 4(a+ b+ c) + 8

=−3− 4(0) + 12

−(−1) + 2(−3)− 4(0) + 8

=9

3= 3


While this method is perfectly fine, it involves quite a bit of computation.It’s extremely easy to make a mistake with lots of expansions in severalvariables, so we have to be very careful when working with these types ofmanipulations. Fortunately, we have a better method.

Solution. Define a completely new polynomial Q(x), which satisfies Q(x) =P (2 − x) for all x. We do this because since the roots of P (x) are a, b andc, the roots of Q(x) are r1 = 2− a, r2 = 2− b, and r3 = 2− c. Furthermore,we can calculate

Q(x) = P (2− x) = (2− x)3 − 3(2− x) + 1 = −x3 + 6x2 − 9x+ 3

Now we are looking for 1r1

+ 1r2

+ 1r3

= r1r2+r2r3+r1r3r1r2r3

, which is a much simplerproposition. This is a direct application of Vieta’s formulas, giving us (again)93

= 3 .

We can even take this one step further! Let’s take a look at a moregeneral case when


n−1 + . . .+ a0

has the roots r1, r2, . . . , rn. We already saw how to find the polynomial withroots, say, 2 − r1, 2 − r2, . . . , 2 − rn. The same holds for polynomials withroots such as 2r1 + 3, 2r2 + 3, . . . , 2rn + 3. But what if we want to findthe polynomial with roots 1

r1, 1r2, . . . , 1

rn? Fortunately, this is essentially the

same method. As P (r1) = P (r2) = . . . = P (rn) = 0, we can (apparently)define a new polynomial Q(x) = P ( 1

x), which has the roots 1

r1, 1r2, . . . , 1

rn.

This would be perfect, except for the fact that Q(x) is not a polynomial! Ifwe plug in 1

xfor P (x), we get

Q(x) =anxn

+an−1xn−1

+ . . .+ a0

But think about what we’re looking at. We’re just looking at the roots,which means that we want Q(something) = 0. In particular, we can multiplyboth sides of the equation by anything convenient. Thus, we could insteadwrite

Q(x) = an + an−1x+ . . .+ a0xn

This would still have the roots 1r1, 1r2, . . . , 1

rn, as all we did is multiply by xn.

Now we can see the relationship between Q(x) and P (x) - their coefficients


are reversed! So we could make this problem even easier, by defining Q1(x) =3x2 − 9x2 + 6x− 1. If this had roots s1, s2, s3, we know that

s1 =1

r1=

1

2− a, s2 =

1

r2=

1

2− b, s3 =

1

r3=

1

2− cand so we are simply looking for s1 + s2 + s3! This is an even more directapplication of Vieta’s formulas, giving us s1 + s2 + s3 = −−9

3= 3.

In this problem, defining new polynomials based on one we already hadserved essentially to reduce our computation. However, it can do so muchmore. First, though, we’ll see a clever application of Vieta’s formulas “inreverse”.

Problem 1.0.7. Real numbers x, y, and z satisfy the equations

x+ y + z = 17

xy + yz + xz = 94

xyz = 168

Find all possible triples (x, y, z).

Solution. At first glance, this hardly seems like a nice problem. A firstinstinct might be to attempt to isolate one variable and solve the resultingequation, but this is ugly, work-intensive, and very easy to make a mistake.Instead we have better methods. Let’s look at the polynomial with roots x,y, and z. Because there are 3 roots, this polynomial is a cubic. Furthermore,we’re given the coefficients (albeit indirectly), as the symmetric sums of{x, y, z} are exactly the equations we’ve received. Thus, we can write

P (a) = (a− x)(a− y)(a− z) = a3 − 17a2 + 94a− 168.

Notice how we can assume P (a) is monic, as the leading coefficient cancelsout anyway.

Now we need to determine the roots of a3 − 17a2 + 94a − 168. Thisis a topic that we haven’t talked about too much, so we’ll go off on a bitof a tangent here. Let’s focus on integer polynomials (those with integercoefficients) for the moment, and do a bit of number theory. Let’s let


n−1 + . . .+ a0

have an integer root r. In other words,

P (r) = anrn + an−1r

n−1 + . . .+ a0 = 0.

Notice that a0 = −anrn − an−1rn−1 − . . .− a1r, which is a multiple of r. Assuch, a0 is also a multiple of r, or more conveniently


Theorem 1.0.8. If P (x) is an integer polynomial, then the only possibleinteger roots of the polynomial are those that divide its constant term.

We can easily extend this concept to rational roots. Let’s say that P (x) hasa rational root r = p

q, where p and q are relatively prime integers. Then we

have

P (r) = an

(pn

qn

)+ an−1

(pn−1

qn−1

)+ . . .+ a0 = 0

=⇒ anpn + an−1p

n−1q + . . .+ a1pqn−1 + a0q

n = 0

Notice that every term except the last is a multiple of p, so we must havep | a0qn. But p and q are relatively prime, so p | a0. Similarly, every termexcept the first is a multiple of q, so we must have q | an. This is the RationalRoot Theorem, more succinctly stated as

Theorem 1.0.9 (Rational Root Theorem). If P (x) is an integer poly-nomial, then the only possible rational roots of the polynomial are ofthe form p

q, where p divides the constant term of P (x) and q divides its

leading coefficient.

On their own, these don’t seem particularly useful. And in a theoreticalsense, they are not. But when trying to do things such as factor a3− 17a2 +94a− 168, we can restrict our search for roots significantly. This won’t helpus when there are non-rational roots, but most of problems that end uprequiring a factorization will have rational roots. Should you come acrossone that does not, there are generally 3 possibilities:

• You made a computational error somewhere

• You don’t actually need the specific factorization. For example, is itenough to know the sum of the roots? Might you only need to knowthat there exists a real root?

• You don’t need to fully factor the polynomial, only divide it into somesmaller parts. For example, you cannot factor x2 + 2 (besides (x −√

2)(x +√

2), which is sometimes useful), but this doesn’t mean youcan’t factor x3 + x2 + 2x+ 2 into (x2 + 2)(x+ 1).

Don’t forget that roots can be negative as well! Also notice that the RationalRoot Theorem (or RRT) both implies the previous theorem and implies that


if a polynomial is monic, the only possible rational roots are also integers.Applying this to our polynomial a3−17a2 + 94a−168, the only numbers weneed to check are those which are factors of 168. It’s not difficult to verifythat a = 4, a = 6, and a = 7 work, so we have

a3 − 17a2 + 94a− 168 = (a− 4)(a− 6)(a− 7)

which implies that (x, y, z) = (4, 6, 7) and permutations.

There is one last method to figuring out the roots of polynomials thatmay prove useful, but we actually need a bit of calculus to rigorously for-mulate it. We’ll allow it to just make “intuitive sense” for now.

Theorem 1.0.10 (Intermediate Value Theorem (IVT)). If P is a poly-nomial (in fact, it suffices for P to be continuous, but because all polyno-mials are continous, we can apply this here) and P (a) = a1, P (b) = b1,for all c1 ∈ [a1, b1] there exists a c ∈ [a, b] such that P (c) = c1.

This seems like quite a mouthful, but in reality the concept is quite simple.If we think about a graph of the polynomial, what this essentially says is thatif we have a line and two points on opposite sides of it, then any continuouscurve between the two points must intersect the line at some point (isn’t itamazing how we require “higher” and more difficult math to prove easierand more obvious things?).

This becomes important when a1 < 0 and b1 > 0. Then we can say thereexists a c such that a < c < b and P (c) = c1. For example, in our previousexample, we had P (a) = a3 − 17a2 + 94a − 168. Since P (0) = −168 andP (10000) is a very big number, this implies that there is some c between 0and 10000 such that P (c) = 0, or c is a root. More usefully, since P (5) = 2,there must be a root between 0 and 5. Although this is, once again, notnecessarily helpful for determining the actual roots themselves, combiningthese usually gives us a great place to start our search. If we come up empty,at least we know that the roots of the polynomial are not rational.

Let’s step away from these “estimation” methods and get back to somereal math. Remember the importance of defining new convenient polynomi-als? Oftentimes we need to do this multiple times to solve our problem, orcome up with a rather non-obvious definition. For example, an extremelyclassic problem runs

Problem 1.0.8. If f(x) is a polynomial of degree 2008 such that f(m) = 1m

for m = 1, 2, 3, . . . , 2009, find f(2010). (Duke Math Meet, 2010)


Solution. We are given 2009 values of a polynomial that has degree 2008, sowe should indeed (theoretically) be able to determine the exact polynomial.This, however, doesn’t tell us anything remotely useful about how to do this.We might try our standard trick of writing the polynomial in terms of itsroots, but we have next to no information about the roots. Let’s digress fora second and think about an easier problem:

Example 1.0.1. Let f(x) be a monic polynomial of degree 2009 suchthat f(m) = 1 for m = 1, 2, 3, . . . , 2009. Find f(2010).

Solution. Here, the solution is significantly more obvious. Let g(x) = f(x)−1, which implies that g(x) has the roots 1, 2, 3, . . . , 2009, meaning thatg(x) = (x − 1)(x − 2) . . . (x − 2009). Thus f(x) = (x − 1)(x − 2) . . . (x −2009) + 1, and so f(2010) = 2009! + 1.

We’d like to find a way to apply this to our problem, but for this we need tocome up with a g(x) that has convenient roots. We are given that f(m) = 1

m

for m = 1, 2, . . . , 2009, but writing this as f(m) − 1m

= 0 is completelyunhelpful. Instead we could maybe write this as mf(m) − 1 = 0, which isindeed a much better idea! Let’s define g(x) = xf(x)− 1. Notice that g(x)has degree 2009, and has the roots 1, 2, 3, . . . , 2009. As such, we can write

g(x) = a(x− 1)(x− 2) . . . (x− 2009)

for some constant a. We’ve made serious progress, but now we appear to beat a roadblock. We’re only given 2009 values, and each of them cause g(x)to evaluate to 0. To figure out a, we’d need one more value. Fortunately,we have it! Recall that

xf(x)− 1 = g(x) = a(x− 1)(x− 2) . . . (x− 2009)

so plugging in x = 0 gives us

0f(0)− 1 = −1 = a(−1)(−2) . . . (−2009) = −2009!a

=⇒ a =1

2009!

Now we can finish the problem by simply plugging in f(2010), which givesus

2010f(2010)− 1 = g(2010) =1

2009!(2009)(2008) . . . (1) = 1


=⇒ 2010f(2010)− 1 = 1 =⇒ f(2010) =1

1005

This type of trick is what makes up many supposedly difficult polynomialproblems. Understanding the importance of roots typically allows us to comeup with the appropriate new polynomial (or polynomials!) to work with.Sometimes this can be difficult, as in the previous problem, when figuringout the appropriate values is not so obvious. Sometimes, as in the nextexample, it’s quite a bit easier (if somewhat longer, and therefore slightlymore difficult to keep track of).

Problem 1.0.9. Suppose P (x) is a polynomial such that P (1) = 1 and

P (2x)

P (x+ 1)= 8− 56

x+ 7

for all real x for which both sides are defined. Find P (−1). (2003 HMMT)

Solution. The first thing we do is manipulate the given equation:

P (2x)

P (x+ 1)= 8− 56

x+ 7=⇒ P (2x)

P (x+ 1)=

8x

x+ 7

=⇒ 8xP (x+ 1) = (x+ 7)P (2x)

There are several “obvious” ways of continuing from here, but they all involveplugging in convenient values. How do we define convenient? Well, primecandidates would be those that cause heavy cancellations, such as x = 0 orx = −7 in our example above. Let’s choose the former, which implies that

8(0)P (1) = (0 + 7)P (0) =⇒ P (0) = 0

which implies that 0 is a root of P (x). In other words, there exists a poly-nomial Q(x) such that P (x) = xQ(x). Let’s plug this in:

8xP (x+ 1) = (x+ 7)P (2x) =⇒ 8x(x+ 1)Q(x+ 1) = 2x(x+ 7)Q(2x)

=⇒ 4(x+ 1)Q(x+ 1) = (x+ 7)Q(2x)

Now we could plug in either x = −1 or x = −7, the former being slightlysimpler to work with. This gives us

0 = 6Q(−2) =⇒ Q(−2) = 0


which implies that there exists an R(x) with Q(x) = (x+ 2)R(x). Pluggingthis in gives us

4(x+ 1)(x+ 3)R(x+ 1) = 2(x+ 1)(x+ 7)R(2x)

=⇒ 2(x+ 3)R(x+ 1) = (x+ 7)R(2x)

Finally, plugging in x = −3 gives us R(−6) = 0, so there exists a polynomialS(x) such that R(x) = (x+ 6)S(x). Plugging this in gives us

2(x+ 3)(x+ 7)S(x) = 2(x+ 3)(x+ 7)S(2x)

=⇒ S(x) = S(2x)

Now we’ve got a bit of an issue. What kind of polynomial could this possiblybe? Here we can use a clever trick. Notice that S(1) must be something- let’s call it c. Thus we have c = S(1) = S(2) = S(4) = . . ., implyingthat the polynomial S1(x) = S(x)− c has roots 1, 2, 4, 8, . . .. A polynomialwith an infinite amount of roots? This has more roots than its degree, andso it must be the 0 polynomial! As such we have S(x) = c for all x, andso S(x) is a constant polynomial. This type of trick is extremely commonto prove two polynomials are equal. In fact, this gives rise to an alternateformulation of the identity theorem:

Theorem 1.0.11 (Identity Theorem). If P (x) and Q(x) are polyno-mials with degree m and n respectively, and P (x) = Q(x) for at leastmax(m,n) + 1 values, then P (x) = Q(x) for all x.

This, of course, makes perfect intuitive sense. If we’re given a polynomialwith degree n, and we know n + 1 of the values, there is exactly one poly-nomial satisfying these conditions. As such, if two polynomials satisfy theseconditions, then they must actually be the same polynomial. Finally, wecan finish the problem. Putting everything together, we have

P (x) = xQ(x) = x(x+ 2)R(x) = x(x+ 2)(x+ 6)S(x)

=⇒ P (1) = 1 = 1(3)(7)S(1) =⇒ S(1) =1

21=⇒ S(x) =

1

21

for all x, as S(x) was a constant polynomial. Finally,

P (x) =1

21x(x+ 2)(x+ 6) =⇒ P (−1) = − 5

21.


This problem was quite long, and we even needed to expand a previoustheorem. However, there was nothing particularly complicated about it.We simply plugged in some convenient values, used the Factor Theoremrepeatedly, and ended up determining our desired polynomial. Certainlythis was much easier than the previous problem, where we need to be quiteclever in figuring out our polynomial!

We’ll look at one more problem to drive our concepts home, and thenconclude this already extremely long article with some practice exercises.

Problem 1.0.10. Let p(x) = x5 + x2 + 1 have roots r1, r2, r3, r4, r5. Letq(x) = x2 − 2. Compute q(r1)q(r2)q(r3)q(r4)q(r5). (USAMTS)

Solution. This may look a bit confusing at first glance - a polynomial withina polynomial? Polynomial-ception? But if we take a moment to take in theproblem statement, we realize we are simply looking for

(r21 − 2)(r22 − 2)(r23 − 2)(r24 − 2)(r25 − 2).

Our first instinct might be “great! Let’s expand this entire expression anduse Vieta’s formulas!”. While a valid interpretation, and one that can indeedbe made to work, this is really quite difficult. This is because the resultingsums of products is of the form (to use one example)

r21r22r

23r

24 + r21r

22r

23r

25 + r21r

22r

24r

25 + r21r

23r

24r

25 + r22r

23r

24r

25

Now the issue becomes apparent. Though Vieta’s formulas tell us lots aboutthe symmetric sums of {r1, r2, r3, r4, r5}, they don’t say anything about thesymmetric sums of {r21, r22, r23, r24, r25}.

Your next instinct, upon recognizing this problem, might be to con-struct the polynomial with roots r21, r

22, . . . , r

25. While an excellent idea, it’s

not quite implementable. Taking Q(x) = P (√x) would seem quite logical,

but Q(x) is not a polynomial and so we can’t use our familiar techniquesanymore.

So we have, once again, hit an apparent roadblock. But, as always, allis not lost. Thinking that it would be quite nice to have (r1 − 2)(r2 −2) . . . (r5 − 2) instead, we can note that just as nice is anything of the form(r1 − c)(r2 − c) . . . (r5 − c) for pretty much any c. This is our “eureka!”moment - we can factor r21 − 2 into (r1−

√2)(r1 +

√2)! Applying this gives

(r21 − 2)(r22 − 2)(r23 − 2)(r24 − 2)(r25 − 2)

= (r1 −√

2)(r2 −√

2) . . . (r5 −√

2)(r1 +√

2)(r2 +√

2) . . . (r5 +√

2).


Now we can use our standard transformation techniques. Let’s let

Q(x) = P (x+√

2) = (x+√

2)5 + (x+√

2)2 + 1

= x5 + 5x4√

2 + 20x3 + (20 +√

2)x2 + (20 + 2√

2)x+ (4√

2 + 3)

Notice that this polynomial has roots r1 −√

2, r2 −√

2, . . . , r5 −√

2. ByVieta’s formulas, their product is

(r1 −√

2)(r2 −√

2) . . . (r5 −√

2) = −4√

2 + 3

1= −(4

√2 + 3).

Now we use a similar process for the other product. Let

R(x) = P (x−√

2) = (x−√

2)5 + (x−√

2)2 + 1

= x5 − 5x4√

2 + 20x3 − (20√

2− 1)x2 + (20− 2√

2)x− (4√

2− 3)

Notice that the roots of this polynomial are r1 +√

2, r2 +√

2, . . . , r5 +√

2,so their product is

(r1 +√

2)(r2 +√

2) . . . (r5 +√

2) = 4√

2− 3

Therefore, the final product is

(r21 − 2)(r22 − 2) . . . (r25 − 2) = −(4√

2 + 3)(4√

2− 3) = −(32− 9) = −23 .

Summary:The field of polynomials (a small joke to those familiar with rings) is

a giant part of all of Algebra. Unsurprisingly, as a result they appear onmathematics contests everywhere. Their accessibility as simple functionsprovides a certain appeal both to the problem solver and the problem writer,and the wide range of techniques available for attacking them means thatthere are many different types of problems to write. Ranging from simplyevaluating a polynomial at certain points (the common “function table”) tocomplicated techniques like the ones we saw in some of the later problems,the methods used to solve polynomial problems vary widlly in both styleand complexity.

Fortunately, much of this polynomial theory is quite natural. We startedoff with some simple explorations, highlighting the similarities between poly-nomials and integers. In fact, the beginning of this summary was slightly


more than a small joke - this type of group theory actually motivates someof this similarity. Moving on, we saw the incredible importance of the rootsof a polynomial, working through simple results such as

r is a root of P (x) ⇐⇒ x− r | P (x)

to significantly more complicated relationships such as Vieta’s formulas. Al-though these alone provide the tools to deal with a great many problems, itis hardly appropriate to simply stop there. We went off on a small tangent,exploring some ways to practically try and evaluate the roots of a polyno-mial by hand, though these methods are really only of practical importancein the world of mathematics competitions where such things are thought ofwhile the problem is being written. Getting back to the theoretical side ofthings, we went through several methods of transforming polynomials, rang-ing from simple transformations such as Q(x) = P (x+ 1) to more complexones such as Q(x) = xP (x) − 1. As we saw, many of the more difficultpolynomial problems rely on these type of substitutions. Finally, our finalproblem illustrated ways for us to apply our methods in somewhat cleverways, taking advantage of factorizations (and sometimes the reverse, expan-sions) to reach simpler polynomials.

Exercises:This is the exercises section. In every article we include around 10 exer-

cises (in this particular article, which is essentially two articles, several more)designed to enhance the reader’s understanding of a topic through their ownstruggles with problems. After learning new techniques, it is practically nec-essary to apply them to problems as soon as possible.

1. When a polynomial p(x) is divided by:

• (x− 2)(x− 3), the remainder is 2

• (x− 1)(x− 3), the remainder is 5− x• (x− 1)(x− 2), the remainder is 6− 2x

What is the least possible degree of p(x)?

2. Define the complex conjugate of a complex number z = a + bi,denoted z, to be a − bi. Notice that the conjugate of a real numberis the real number itself. Prove that z + w = z + w and zw = z · w,


where w and z are complex numbers. Use this fact to prove that if zis a root of P (x), then so is z.

3. Pick an arbitrary polynomial P (x) with integer coefficients (to makeyour life easier, keep the coefficients reasonable). Compute each of thefollowing numbers:

P (3)− P (1), P (4)− P (2), P (5)− P (3), P (6)− P (4)

What do these numbers have in common? What about

P (4)− P (1), P (5)− P (2), P (6)− P (3)?

Can you generalize this observation to P (a)− P (b)?

4. Let P (x) be a polynomial with integer coefficients. Show that if P (0)and P (1) are both odd numbers, then P (x) cannot have any integerroots. (Canada)

5. Let P (x) be a polynomial with integer coefficients. If P (5) = 25, P (14) =16, and P (16) = 36, find the sum of all possible 3-digit values of P (10).

6. Suppose that the roots of x3 + 3x2 + 4x− 11 = 0 are a, b, and c, andthat the roots of x3 + rx2 + sx+ t = 0 are a+ b, b+ c, and c+ a. Findt. (1996 AIME)

7. Find all solutions to the system

x+ y + z = 2

x2 + y2 + z2 = 14

xyz = −6.

8. Find b if the equation x4 + ax3 + bx2 + cx+ d has rational coefficientsand 4 complex roots, two with sum 3 + 4i and the other two withproduct 13 + i. (1995 AIME)

9. The product of two of the four roots of the quartic equation

x4 − 18x3 + kx2 + 200x− 1984 = 0

is −32. Determine the value of k. (1984 USAMO)


10. Find all integer polynomials P (x) such that

xP (x− 1) = (x− 26)P (x)

holds for all x.

11. Find all polynomials P (x) such that 2P (x) = P (x− 1) + P (x+ 1).

12. Given

P (x) = (1 + x+ x2)1000 = a0 + a1x+ a2x2 + . . .+ a2000x

2000,

compute S = a0 + a2 + . . .+ a2000.

13. Let P (x) be a polynomial of degree 10 such that P (2i) = i for 0 ≤ i ≤10. What is the coefficient of x1 in P (x)?

14. The complex numbers α1, α2, α3, α4 are the 4 distinct roots of x4+2x3+2 = 0. Determine the unordered set {α1α2 +α3α4, α1α3 +α2α4, α1α4 +α2α3}. (2007 HMMT)

15. Find the coefficient of x2 in the polynomial

(. . . (((x− 2)2 − 2)2 − 2)2 . . .− 2)2,

where there are n pairs of brackets.

16. Kelvin the frog enjoys generating random matrices of numbers. Hehas a very particular method for doing so. First he chooses 4028real numbers a1, a2, . . . , a2014, b1, b2, . . . , b2014, then he draws a 2014 ×2014 grid of unit squares. On top of this grid he writes the numbersa1, a2, . . . , a2014, one per column, and along the left side of the gridhe writes the numbers b1, b2, . . . , b2014. In each square, he writes thesum of the number in its column and the number in its row (so, forexample, the number in the 5th column and the 2nd row would bea5 + b2). Let r1, r2, . . . , r2014 be the product of the numbers in the 1st,2nd, . . ., 2014th rows respectively, and c1, c2, . . . , c2014 be the productof the numbers in the 1st, 2nd, . . ., 2014th columns respectively. Oneday Kelvin the frog notices that his chosen matrix satisfies r1 = r2 =. . . = r2014 = 2014. Compute c1.


17. Let P (n) be a polynomial of degree m with integer coefficients, wherem ≤ 10. Suppose that P (0) = 0, P (n) hasm distinct integer roots, andP (n)+1 can be factored as the product of two nonconstant polynomialswith integer coefficients. Find the sum of all possible values of P (2).(NIMO)

18. Show that there does not exist a non-constant polynomial P (x) suchthat [P (x)]2 − 1 = P (x2 + 1) for all real x.

19. Let a, b, c be three distinct integers, and let n be a positive integer.Show that

an

(a− b)(a− c)+

bn

(b− a)(b− c)+

cn

(c− a)(c− b)

is an integer (Hungary 1959)

20. Find all polynomials P (x) with real coefficients such that

P (a− b) + P (b− c) + P (c− a) = 2P (a+ b+ c)

for all real numbers a, b, c such that ab+ bc+ ca = 0.

2

AIME Combinatorics Part I

Combinatorics, as it extends to the AIME, is somewhat difficult to quantifyinto specific broad areas (unlike Algebra, where we can break the subjectdown into things like polynomials, functional equations, inequalities, andso on, or Geometry where we can talk about cyclic quadrilaterals, trianglegeometry, trignometry, and so on and so forth). There are, more or less(and with respect to the AIME), two different types of combinatorics (again,besides recursion):

1. Those that are simply “do it”. In these types of problems, you simplydetermine the valid cases and do it out.

2. Those that involve clever manipulations or observations to reach theprevious point.

Of course, we can further subdivide these categories to include things suchas “conditional probability” and “PIE” and all sorts of other formulae andtheorems, but this rather broad (and arbitrary) definition is sufficient fornow.

Central to our investigation is the binomial coefficient. This article fo-cuses on binomial identities, which loosely refers to indentities expressiblein binomial coefficients. As we shall see, there are algebraic and combina-torical ways to prove these identities, the latter typically being “nicer” andmore motivated. Written as

(nk

), this denotes the number of ways to select

k objects from a set of n objects, without respect to order. We can think ofthis as follows. There are n ways to select the first object, n−1 ways to selectthe second object, and so on up to n− k + 1 ways to select the kth object.However, we are drastically overcounting. Choosing, for example, “ABC” is

25


effectively the same thing as choosing “BAC” (when order is disregarded),or “CBA”, and so on. Since there are k! = k · (k− 1) · (k− 2) · . . . · 2 · 1 waysto arrange our set of k objects, this is what we need to divide by. This givesus a convenient formula(

n

k

)=n · (n− 1) · (n− 2) · . . . · (n− k + 2) · (n− k + 1)

k!=

n!

k!(n− k)!

Notice that this is only valid when n ≥ k. Otherwise, we resort to ourcombinatorical definition. If k > n, we are asking for the number of waysto choose a k-person committee from a set of n < k people, which is clearlyimpossible. Thus we define

(nk

)to be 0 when n < k. Notice that this

immediately gives us our first binomial identity(n

k

)=

n!

k!(n− k)!=

n!

(n− k)!(n− (n− k))!=

(n

n− k

)This formula, while innocuous at first glance, allows us to motivate severalidentities by using double-counting arguments. In these arguments, wecount something in two different ways. Because they count the same thing,they must be equal. In fact, we can apply this concept to our first binomialidentity above. Notice that choosing k objects from a group of n is exactlythe same as choosing n− k obejcts to not be in our group, so indeed

(nk

)=(

nn−k

)as expected.

As another simple example, let’s count the number of ways to choosea 2-person committee from a set of n + 1 people. We already know thatthis is

(n+12

), but we can also count it in a different way. Call the peo-

ple 1, 2, 3, . . . , n + 1. If person 1 is in the committee, then we can chooseany of the remaining n people to serve. Otherwise, if person two is in thecommittee, we can choose n − 1 of the remaining people (since we alreadyestablished person 1 is not in the committee) to serve. Continuing in thisfashion, there are n + (n − 1) + (n − 2) + . . . + 3 + 2 + 1 ways to form thecommittee. As such, we can conclude

1 + 2 + 3 + . . .+ (n− 1) + n =

(n+ 1

2

)=

(n+ 1)!

(n− 1)!2!=n(n+ 1)

2

which is a very familiar formula, which has a myriad of other proofs (alge-braic, inductive, and so on).

These type of arguments are sometimes also called committee-formingarguments. The reason for this is obvious - we are counting the number


of ways to form a committee in two different ways. Another simple examplestems from choosing a k person committee from a group of n people. Wealready defined this as

(nk

), but we can also count it in another (recursive)

way. Consider any person in the group A. If this person is in the group, nowwe need to choose a k − 1 person committee from a group of n− 1 people,which there are

(n−1k−1

)ways to do. If this person is not in the group, now we

need to choose a k person committee from a group of n − 1 people, whichthere are

(n−1k

)ways to do. This gives us the relation

Theorem 2.0.12 (Pascal’s Identity).(n

k

)=

(n− 1

k

)+

(n− 1

k − 1

)

Although a combinatorical argument was practically necessary to come upwith this relation, once we have this relation on paper and we want to verifyit, we can take an algebraic approach:

Proof. Simply rewrite the binomial coefficients in their factorial forms, giv-ing us (

n

k

)=

(n− 1

k

)+

(n− 1

k − 1

)

⇐⇒ n!

k!(n− k)!=

(n− 1)!

k!(n− k − 1)!+

(n− 1)!

(k − 1)!(n− k)!

Multiplying through by k!(n− k)! gives us

⇐⇒ n! = (n− 1)!(n− k) + (n− 1)!(k) = (n− 1)!(n− k + k) = (n− 1)!(n)

which is indeed clearly true.

This identity forms the basis for Pascal’s triangle. This triangle begins


1

1 1

1 2 1

1 3 3 1

1 4 6 4 1

1 5 10 10 5 1

1 6 15 20 15 6 1

1 7 21 35 35 21 7 1

The top row, or the zeroth row, of Pascal’s triangle is simply a 1, andthe next row (the first row) consists of two 1s. Every number after thatis the sum of the two numbers directly above it. Notice that the first rowconsists of

(10

)and

(11

), so in fact every number in this triangle is a binomial

coefficient! In particular, the kth number in the nth row (starting fromk = 0) is

(nk

).

(00

)(10

) (11

)(20

) (21

) (22

)(30

) (31

) (32

) (33

)(40

) (41

) (42

) (43

) (44

)(50

) (51

) (52

) (53

) (54

) (55

)(60

) (61

) (62

) (63

) (64

) (65

) (66

)(70

) (71

) (72

) (73

) (74

) (75

) (76

) (77

)


Several binomial identities can be likened to paths on Pascal’s triangle.For example, if we repeatedly apply Pascal’s Theorem, we get(

n

k

)=

(n− 1

k

)+

(n− 1

k − 1

)=

(n− 2

k

)+

(n− 2

k − 1

)+

(n− 1

k − 1

)

=

(n− 3

k

)+

(n− 3

k − 1

)+

(n− 2

k − 1

)+

(n− 1

k − 1

)= . . .

Repeating this process kown to(k+1k

)=(kk

)+(kk−1

), we get(

n

k

)=

(k

k

)+

(k

k − 1

)+

(k + 1

k − 1

)+ . . .+

(n− 2

k − 1

)+

(n− 1

k − 1

)Notice that

(kk

)= 1 =

(k−1k−1

), so we can write this in the nicer form(

k − 1

k − 1

)+

(k

k − 1

)+ . . .+

(n− 2

k − 1

)+

(n− 1

k − 1

)=

(n

k

)If we adjust the indices to get a nicer form, we have the convenient HockeyStick Identity:

Theorem 2.0.13 (Hockey Stick Identity). For positive integers n ≥ k,we have (

k

k

)+

(k + 1

k

)+ . . .+

(n− 1

k

)+

(n

k

)=

(n+ 1

k + 1

)Unsurprisingly, there is a simple combinatorical interpretation of this iden-tity as well. If we need to choose a committee of size k + 1 from a set ofn+ 1 people, we can list the n+ 1 people as 1, 2, 3, . . . , n+ 1 and divide thisinto the cases

• Case 1: Person 1 is in the committee. Then there are(nk

)ways to

assign the remaining members.

• Case 2: Person 1 is not in the committee, but person 2 is. Then thereare

(n−1k

)ways to assign the remaining members.

• Case 3: Persons 1 and 2 are not in the committee, but person 3 is.Then are are

(n−2k

)ways to assign the remaining members.


• . . .

• Case n − k + 1: People 1, 2, . . . , n − k are not in the committee, butperson n − k + 1 is. Then there are

(kk

)= 1 ways to assign the

remaining members, unsurprising as we are essentially forced to takeall the remaining members (which there is 1 way to do)

Almost every binomial indentity has such a combinatorical interpreta-tion, as well as a relatively clean algebraic method. We can even approachsuch identities with induction, where we most likely use Pascal’s identityin some form. Oftentimes a single identity will have many valid avenues ofattack. For example, let’s take the simple question

Example 2.0.2. Find the sum(n

0

)+

(n

1

)+ . . .+

(n

n− 1

)+

(n

n

)in terms of n.

Solution. [Double Counting] Let’s think about what this sum represents.Given a group of n people, the first term is the number of ways to selecta committee of zero people. The second is the number of ways to selecta committee of 1 person. In general, the kth term in the number of waysto select a committee of k people. Because k runs from 1 to n, the sumrepresents the number of ways to select any committee from a group of npeople. But this is easy - each of the n people can either be in the group ornot, and so there are 2n possibilities.

This has the advantage of being quick, clean, and completely devoid ofpotential for computational mistakes. However, this type of insight is notnecessarily the best way to go about a problem under contest conditions.In those cases it is usually better to discover a pattern and then prove it insome fashion. For example,

Solution. The first thing we do is calculate the sum for a couple small valuesof n. When n = 1, we have

(10

)+(11

)= 1 + 1 = 2, when n = 2 we have(

20

)+(21

)+(22

)= 1 + 2 + 1 = 4, when n = 3 we have

(30

)+(31

)+(32

)+(33

)=

1 + 3 + 3 + 1 = 8, and so on. We hypothesize that the sum is 2n.To prove this, we proceed by induction. The base cases are already

established. Assume that there exists a k such that


(k

0

)+

(k

1

)+ . . .+

(k

k

)= 2k.

Notice that (k + 1

0

)+

(k + 1

1

)+ . . .+

(k + 1

k + 1

)= 1 +

(k

1

)+

(k

0

)+

(k

2

)+

(k

1

)+ . . .+

(k

k

)+

(k

k − 1

)+ 1

Replacing 1 with(k0

)and

(kk

)respectively, we get

= 2

((k

0

)+

(k

1

)+ . . .+

(k

k

))= 2(2k) = 2k+1

Completing the induction. Therefore the answer is 2n for all n.

While it appears on the surface that the latter argument is longer andmore involved, in reality it does not take very much effort at all to comeup with. Checking small values to discover a general pattern is a standardtechnique, and then working out the induction is often a matter of relativelyminor details. The former argument is certainly the nicer one, but pointsare awarded for correctness and not elegence.

Finally, the most general method to solve this problem involves the Bi-nomial Theorem.

Theorem 2.0.14. For positive integers n we have

(x+y)n =

(n

0

)xn+

(n

1

)xn−1y+

(n

2

)xn−2y2+. . .+

(n

n− 1

)xyn−1+

(n

n

)yn.

The proof of this is a relatively straightforward induction, but under-standing why it is true is significantly less involved. If we write the productas

(x+ y)(x+ y)(x+ y) . . . (x+ y)

then it is clear that any term is comprised of n variables, each of which is xor y. To form a term of the form xkyn−k, we must choose x exactly k timesfrom a group of n, which we already know there are

(nk

)ways to do.


Applying this to our problem, the binomial coefficients are already present.We simply need all the xkyn−k terms to be 1, occurring when x = y = 1.This gives us

(1 + 1)n =

(n

0

)+

(n

1

)+ . . .+

(n

n

)= 2n.

The Binomial Theorem has a wide range of implications, immediatelygiving us several useful sums. For example, we can instantly get

(1− 1)n =

(n

0

)−(n

1

)+

(n

2

)− . . .+ (−1)n

(n

n

)= 0

This also has a combinatorical interpretation. We are searching for thedifference between the number of committees with an even number of peopleand the number of committees with an odd number of people, each formedfrom a group of n people. The result is implying that these are equal, whichwe can in fact quickly see combinatorically. If we arbitrarily assign n− 1 ofthe people to either be in the group or not, which there are 2n−1 ways to do,then the final person has exactly two choices - be in the group, or don’t bein the group. Exactly one of these choices will result in a committee withan even number of people, and the other will result in a committee with anodd number of people. Thus, the difference is 0.

Notice how we proved as an intermediate step that(n

0

)+

(n

2

)+ . . .+

(n

2bn2c

)= 2n−1

which is also evident by examining (1+1)n+(1−1)n2

= 2n−1 with the sumsthat we found above.

Our final example of double counting involves what is known as the“chairperson” or “subcommittee” method. Usually used when dealing withproducts of binomial coefficients, the crux of the argument is that there arethe same number of ways to choose a committee and then a chairperson(or subcommittee) of this committee as there are ways to choose a chair-person (or subcommittee) and then the remaining committee members. Forexample,


Example 2.0.3. Show that

k

(n

k

)= n

(n− 1

k − 1

)Solution. There is, of course, a straightforward algebraic solution:

k

(n

k

)= k· n!

k!(n− k)!=

n!

(n− k)!(k − 1)!= n· (n− 1)!

(n− k)!(k − 1)!= n

(n− 1

k − 1

)but we can use our king argument in a straightforward manner as well.Notice that k

(nk

)represents the number of ways to choose a k-person com-

mittee, and then a chairperson from among these k people. But we can alsorepresent this as the number of ways to choose a chairperson from the entiregroup (which is n) times the number of ways to choose k − 1 people fromthe remaining n− 1, or

(n−1k−1

). Thus k

(nk

)= n

(n−1k−1

)as desired.

There are, unsurprisingly, a great deal of arguments of this sort. In fact,a quite reasonable way to create problems related to binomial identities isby choosing any 2 equivalent ways of determining subcommittees and ask-ing to show that their binomial representations are equal. As such, goingthrough many more examples is rather counterproductive when the funda-mentals have already been established. Therefore we should conclude thearticle here and move on to the exercises portion of the chapter.

Summary:Combinatorics, in its simplest form, involves choosing items from a larger

set of the same items. Not surprisingly, this can get complicated. The ter-minology

(nk

)is extremely important, representing the number of ways to

choose k objects from a set of n objects. This quickly gives us a simplealgebraic representation, but this representation often does not lend itself tosimple manipulation. In these cases, we need to step back from the variablesand analyze the combinatorical situation at hand. Most of our combinator-ical arguments are variations of the standard committee-forming argu-ment, looking literally at the situation where we are choosing a committeeof k people from a group of n people. Being able to count this in multipleways motivates many identities that would otherwise prove difficult to workwith.

Exercises:


1. Show that (b

a

)(c

b

)=

(c

a

)(c− ab− a

)both algebraically and combinatorically.

2. Show that kn

(nk

)=(n−1k−1

). What does this mean combinatorically?

3. Show that∑n

i=0(−1)i(ni

)= 0 for all integers n - e.g. prove that

(n0

)−(

n1

)+(n2

)− . . .+ (−1)n

(nn

)= 0.

4. Prove that∑n

i=0(−1)i+1i(ni

)= 0 for all integers n. In other words,

show that

−0

(n

0

)+ 1

(n

1

)− 2

(n

2

)+ . . .+ (−1)n+1n

(n

n

)= 0.

5. Let r, s and n be integers. The next couple of exercises will determinea useful formula. Show that

r+s∑i=0

(r + s

n

)xn = (1 + x)r+s.

6. Use this to show that

r+s∑i=0

(r + s

n

)=

r∑j=0

(r

j

)xj∑k=0

(s

k

)xk.

7. Combining these, show that

r+s∑i=0

(r + s

n

)=

r+s∑i=0

(∑j

(r

j

)(s

n− j

))xi

and find a way to conclude that∑k

(r

k

)(s

n− k

)=

(r + s

n

)for all integers r, s, n. Find a combinatorical explanation for why thisis true.


8. Say there is a grid of unit squares labeled with the usual coordinates(so (0,0) is the bottom left) of size n × k (so the upper right corneris labeled (n, k)). On any turn, we may move either one square tothe right or one square up. Find a formula for the number of pathsfrom (0,0) to (n, k). As a slightly more difficult example, say we haven = 7 and k = 5, but there is a flesh-eating monster at (4, 3). Howmany paths, following the same rules as above, go from (0,0) to (7,5)without the inconvenience of having flesh torn off?

9. Show that, for all integers n, we have

n∑i=0

(n

i

)2

=

(2n

n

).

10. Try to come up with a combinatorical explanation for the above. Thisisn’t so simple! As a hint, recall the 8th exercise.

3

AIME Combinatorics Part II

AIME combinatorics problems are difficult to classify into non-broad cate-gories. We have already mentioned recursion as a large area, but most ofthe other Combinatorics problems are of the “just do it” variety - e.g. thosein which the solution is to simply carefully go through the casework andcompute the answer directly.

However, some knowledge of counting arguments often allows us to solvethese types of questions either more directly or with less computation. Forexample, let’s look at a very simple question:

Problem 3.0.11. How many nonnegative integers a and b satisfy the equa-tion a+ b = 10?

Solution. This is hardly a difficult problem. a can be any of the integers from0 to 10 (inclusive), and b is simply 10− a. Thus there are 11 solutions.

That wasn’t too bad. We can, of course, easily generalize to a + b = nfor some constant n. By the exact same logic as the above, there are n + 1solutions. Let’s extend this a bit:

Problem 3.0.12. How many nonnegative integers a, b, and c satisfy theequation a+ b+ c = 10?

Solution. This one is slightly trickier, but we can still easily just bash outthe cases. If a = 0, then b + c = 10. Then 0 ≤ b ≤ 10 like above, sothere are 11 solutions. If a = 1, then b + c = 9. By similar reasoning,there are 10 solutions in this case. Continuing this logic down to a = 10,then b + c = 0 has just one solution b = c = 0. Thus there are a total of11 + 10 + 9 + . . .+ 2 + 1 = 66 solutions.

36


Let’s try and generalize this one as well:

Problem 3.0.13. How many nonnegative integers a, b, c satisfy the equationa+ b+ c = n?

Solution. As before we list out some cases. When a = 0 we have b+ c = n,which we already established as having n + 1 solutions. When a = 1 wehave b + c = n− 1, which contributes another n solutions. Continuing thisline of reasoning, there are a total of (n + 1) + n + . . . + 2 + 1 = (n+1)(n+2)

2

solutions.

So what if we wanted to extend this to 4 variables a, b, c and d? For-tunately, this does so quite naturally if we write (n+1)(n+2)

2as(n+22

). Then

when we have a+ b+ c+ d = n, looking at the cases a = 0, a = 1, . . . , a = nthere are a total of

(n+22

)+(n+12

)+ . . . +

(22

)solutions. But by the Hockey

Stick identity from the previous article, we know that this is equivalent to(n+33

). This shouldn’t seem strange - this is essentially how we derived the

a+b+c = n case. Notice that (n+1)+n+ . . .+2+1 = (n+1)(n+2)2

is exactly

the same as saying(n+11

)+(n1

)+ . . .+

(11

)=(n+22

), exactly what the Hockey

Stick identity tells us.

If we extend this to even more variables, say a1, a2, . . . , ak, then we canhypothesize that there are

(n+k−1k−1

)solutions by our expressions above. Let’s

prove this in, as usual, two different ways.

Problem 3.0.14. How many nonnegative integers a1, a2, . . . , ak satisfy theequation a1 + a2 + . . .+ ak = n?

Solution. We claim that the answer is(n+k−1k−1

). We proceed by induction.

The base cases when n = 2, 3, 4 were already done above, though we onlyneed one. Assume that the result holds for some k, e.g. that a1+. . .+ak = nhas

(n+k−1k−1

)solutions. We show that a1 + a2 + . . . + ak+1 = n has

(n+kk

)solutions.

Notice that a2 + a3 + . . . + ak+1 = n − a1. The number of solutions tothis equation, by the inductive hypothesis, is

(n−a1+k−1

k−1

). So we are looking

forn∑i=0

(n+ k − 1 + i

k − 1

)which is

(n+kk

)by the Hockey Stick identity as desired.


This proof is very straightforward, and certainly enough to assure usthat our formula is correct. However, it doesn’t really tell us why it’s cor-rect. Seeing that the answer is a simple binomial coefficient suggests thatthere should be a natural combinatorical interpretation of this equation, andindeed there is!

Solution. This method is usually referred to as the “balls and urns” or the“stars and bars” method, the latter likely rising to prominence because of itsease in typing examples on a computer. Consider arranging a set of n+k−1objects, some n of which (the stars) are identical and the remaining k − 1(the bars) are also identical, but different from the first object. For example,if n = 7 and k = 4, then an arrangement such as “**—***—*—*” wouldbe perfectly valid.

Notice that the k−1 bars divide this string into k sections, each of whichcontains some nonnegative amount of stars. If we represent the number ofstars in these sections by a1, a2, . . . , ak, then we have a1 + a2 + . . .+ ak = n- just the equation we were looking for! As such, any arrangement of thatstring leads to a unique solution (and vice versa; this is a bijection betweenthe strings and the solutions to our equation). But finding the number ofarrangements is easy - this is just

(n+k−1k−1

), exactly the formula we were

looking for.

We can, of course, easily extend this to cases where the variables arenecessarily positive as well. In the general case, we want to solve a1 + a2 +. . .+ ak = n with the restriction that a1, a2, . . . , ak are positive integers.

Problem 3.0.15. How many positive integers a1, a2, . . . , ak satisfy theequation a1 + a2 + . . .+ ak = n?

Solution. We can easily transform this into something we know how to solve.Let a1 = b1 + 1, a2 = b2 + 1, . . . , ak = bk + 1. Then this gives us b1 + b2 +. . .+ bk = n− k, where each of the bi are nonnegative integers. We alreadyknow how to do this problem - the answer is

(n−1k−1

).

Combinatorically, this is the same interpretation as before, but we firstput one star in every section to ensure that the variables are positive. Thisleaves us n− 1 objects, k − 1 of which are bars, leading to the result.

Using a bit more cleverness, we can extend this to situations other thandirect equality:

Problem 3.0.16. How many ordered 4-tuples of nonnegative integers (a, b, c, d)satisfy a+ b+ c+ d ≤ 15? (2014 Lehigh Math Competition Problem 32)


Solution. There are two approaches to this problem. We could first use theHockey Stick identity, noting that a + b + c + d = 15 has

(183

)solutions,

a + b + c + d = 14 has(173

)solutions, and so on. This makes the answer(

183

)+(173

)+. . .+

(33

)=(194

). But seeing how the answer is so nice, there should

be a better way of doing things. Indeed, we can introduce a (nonnegativeinteger) variable e so that a + b + c + d + e = 15, making the answer onceagain

(194

).

Well, that seemed to come out of nowhere! Let’s explore why this makesmore sense in the general case:

Problem 3.0.17. How many nonnegative integers a1, a2, . . . , ak satisfy theinequality a1 + a2 + . . .+ ak ≤ n?

Solution. We could, of course, use the Hockey Stick identity once again,checking the cases where a1 +a2 + . . .+ak is n, n−1, n−2, and so on downto 0. This would give us

(n+ k − 1

k − 1

)+

((n− 1) + k − 1

k − 1

)+ . . .+

(k − 1

k − 1

)=

(n+ k

k

)but since this, once again, results in a single binomial coefficient, there

should be a natural interpretation of this result. Unsurprisingly, there is.Notice how we are checking when a1 + a2 + . . .+ ak is all integers less thanor equal to n. As such we can write

a1 + a2 + . . .+ ak = n− x

where x is a nonnegative integer. This rearranges to

a1 + a2 + . . .+ ak + x = n

so now we have the sum of k+1 nonnegative integers equal to n. We alreadyknow how to handle this problem; the answer is

(n+(k+1)−1(k+1)−1

)=(n+kk

).

Combinitorically, this is exactly equivalent to simply adding a bar to ouralready constructed string.

With all of these counting tools, we’ll dive into some recent AIME com-binatorics problems. The first one is a rather simple example - we just needa very simple observation to reduce it to a direct binomial coefficient.


Problem 3.0.18. Let B be the set of all binary integers that can be writtenusing exactly 5 zeros and 8 ones where leading zeros are allowed. If allpossible subtractions are performed in which one element of B is subtractedfrom another, find the number of times the answer 1 is obtained. (2012AIME I Problem 5)

Solution. It’s fairly easy to see that the only way we can end up with a1 result is if we are subtracting the numbers (stuff)10 and (stuff)01. The“stuff” is comprised of 4 zeros and 7 ones, where leading zeros are allowed.Thus, our answer is simply the number of arrangements of 4 zeros and 7ones, or

(4+74

)= 330.

The next problem is slightly harder, but follows the same line of reason-ing.

Problem 3.0.19. Let S be the increasing sequence of positive integerswhose binary representation has exactly 8 ones. Let N be the 1000th num-ber in S. Find the remainder when N is divided by 1000. (2012 AIME IIProblem 7)

Solution. The sequence starts with the numbers with 8 ones and no zeros,continues on to the numbers with 8 ones and 1 zero, and so on. If we allowleading zeros, there are

(n+88

)numbers with at most n zeros, so the 1000th

number in this sequence is the 1000−(128

)= 505th number with 8 ones and

5 zeros.There are

(114

)= 330 numbers that start with 10 in binary, so we are

looking for the 175th number that starts with 11. There are(104

)= 210

numbers beginning with 110, so the next digit of the 1000th number is azero. Continuing on, making the next digit in our number 0 if there aremore numbers that have a zero in the next place than our current “counter”and 1 otherwise, we get that the 1000th number is 11011011110002 = 7032,making our answer 32.

Notice how in the stars and bars approach to problems, we focused onthe “gaps” between the bars as methods for counting. The next problemfollows this same line of reasoning, albeit in a slightly more complex fashion.

Problem 3.0.20. Six men and some number of women stand in a line inrandom order. Let p be the probability that a group of at least four menstand together in the line, given that every man stands next to at least oneother man. Find the least number of women in the line such that p does notexceed 1 percent.


Solution. Let’s say there are x women. Assume all the men are indistin-guishable and all the women are indistinguishable. This is valid because weare simply multipying by a factor of x!6!

x!6!. We can think of placing the men

between the women (or at the front or end), so that there are x+ 1 groups.Because every man stands next to at least one other man, we have 4 cases:

• Case 1: We have 3 non-adjacent groups of 2 men to be placed amongthe women. Since there are x + 1 slots to place 3 such groups, thereare

(x+13

)possibilities in this case.

• Case 2: We have 2 non-adjacent groups of 3 men to be placed amongthe women. Following similar logic to the above case, there are

(x+12

)possibilities.

• Case 3: We have a group of 4 men and a group of 2 men. Now thereare 2

(x+12

)possibilities, as we first choose 2 slots to place the groups

in and then multiply by 2 as our groups are distinguishable.

• Case 4: All 6 men are in a single group. Clearly there are x + 1possibilites in this case.

Only in cases 3 and 4 are 4 men standing next to each other. Thus, theprobability of this happening is

x+ 1 + 2(x+12

)x+ 1 + 2

(x+12

)+(x+12

)+(x+13

) =(x+ 1)2

(x+ 1)(1 + 3x2

+ x2−x6

)=

6x+ 6

x2 + 8x+ 6

which we require to be at most 1%. Thus,

6x+ 6

x2 + 8x+ 6≤ 1

100=⇒ 600x+ 600 ≤ x2 + 8x+ 6 =⇒ 592 +

594

x≤ x.

Therefore, our answer is x = 594 .

Notice how we slightly skated over the conditional probability portionof the solution, because it was quite natural. However, oftentimes problemwriters will test conditional probability as it can be tricky to navigate. Onesimple question of this sort is

Problem 3.0.21. Kelvin the Frog flips 4 coins. Given that at least 2 of themturned up heads, what is the probability that they all turned up heads?


Solution. This one is simple enough. There are precisely(42

)+(43

)+(44

)= 11

possible scenarios where at least 2 coins turned up heads, and each of themare equally likely. There is only 1 case where they are all heads, so ouranswer is 1

11.

The above problem reveals the general strategy for approaching thesetypes of problems. If we want to find the probability of some event A, butwe are also given that event B has occurred, we first find the probability thatevent B occurs (in the above example, 11

16), then calculate the probability that

both A and B occur (in the above example, 116

), then take their ratio. Infancy symbols,

P (A|B) =P (A ∪B)

P (B)

which states, in words, “the probability of A given B is the quotient ofthe probability of A and B occuring and the probability of B occurring”.Notice that this formula assumes P (B) > 0, which is obvious as otherwisethe question is hardly interesting (the answer would simply also be zero).As a slightly more difficult example,

Problem 3.0.22. Kelvin the Frog rolls two dice, one of which is red and oneof which is blue. Given that their sum is at most 6, what is the probabilitythat the red die is showing a 2?

Solution. In this case, the event A is the red die showing a two, and theevent B is the two dice summing to at most 6. The probability of B is easy- there are 15 “good” outcomes out of 36 total, so the probability is 5

12. Now

we need to calculate the probability of A and B occurring. If the red dieis showing a two, then the blue die could be showing a 1, 2, 3, or 4, so theprobability of this is 4

36= 1

9. Thus, the probability that the red die shows a

two given that the sum of the dice is at most 6 is19512

= 415

.

We could go through many many more problems, and still not be ableto come up with a specific focus of attack. As such, the exercises in thisarticle are intentionally broad, spanning several supposed “areas” of combi-natorics. In reality, they are all more or less the same general concept. Weneed to set up our cases carefully (or, sometimes, the lack of cases), and in-terpret scenarios as targets for binomial coefficients. This alone is sufficientto tackle most AIME-level combinatorics problems, but is much easier saidthan done. Missing a single case is disasterous, and miscounting in one is


equally bad. Keeping organized work is essential, as is constant implemen-tation of sanity checks (if I plug in numbers x and y, am I getting what Ishould be getting?) and often listing out cases that you don’t want to makesure you’re covering everything (e.g. if there are four possible cases, andyou want to know the probability that some event falls into one of the firsttwo, instead of just calculating the values for the first two cases you shoulddo so for all four cases and make sure your sum agrees with your generalcase). Combinatorics draws a lot of hate due to these types of silly errors,which ironically leads to students not focusing on combinatorics and thuscommitting even more silly errors. While frustrating, especially on earlyquestions, silly mistakes are often the result of a lack of understanding - nota lack of calculation ability. But both are fixable (and trainable)!

Summary:The area of Combinatorics is very broad, and as such even pinning down

a specific target for attack is incredibly difficult. In this article, we examinedthe famous stars-and-bars method, also commonly known as balls andurns. We also not-so-briefly touched on the notion of conditional probabil-ity, an area that is rather simple mathematically, yet of incredible impor-tance and seemingly counterintuitive. The important formula to rememberis

P (A|B) =P (A ∪B)

P (B)

or “the probability of A given B is the quotient of the probability of A andB occuring and the probability of B occurring”.

Exercises:

1. Dave arrives at an airport which has twelve gates arranged in a straightline with exactly 100 feet between adjacent gates. His departure gateis assigned at random. After waiting at that gate, Dave is told thedeparture gate has been changed to a different gate, again at random.Let the probability that Dave walks 400 feet or less to the new gatebe a fraction m

n, where m and n are relatively prime positive integers.

Find m+ n. (2010 AIME II Problem 4)

2. Find the number of five-digit positive integers, n, that satisfy the fol-lowing conditions:

• the number n is divisible by 5,


• the first and last digits of n are equal, and

• the sum of the digits of n is divisible by 5.

(2013 AIME I Problem 2)

3. Nine people sit down for dinner where there are three choices of meals.Three people order the beef meal, three order the chicken meal, andthree order the fish meal. The waiter serves the nine meals in randomorder. Find the number of ways in which the waiter could serve themeal types to the nine people such that exactly one person receivesthe type of meal ordered by that person. (2012 AIME I Problem 3)

4. Melinda has three empty boxes and 12 textbooks, three of which aremathematics textbooks. One box will hold any three of her textbooks,one will hold any four of her textbooks, and one will hold any fiveof her textbooks. If Melinda packs her textbooks into these boxes inrandom order, the probability that all three mathematics textbooksend up in the same box can be written as m

n, where m and n Are

relatively prime positive integers. Find m+n. (2013 AIME I Problem6)

5. The vertices of a regular nonagon (9-sided polygon) are to be labeledwith the digits 1 through 9 in such a way that the sum of the numberson every three consecutive vertices is a multiple of 3. Two accept-able arrangements are considered to be indistinguishable if one can beobtained from the other by rotating the nonagon in the plane. Findthe number of distinguishable acceptable arrangements. (2011 AIMEI Problem 5)

6. Let N be the number of ordered pairs of nonempty sets A and B thathave the following properties:

• A ∪ B = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12},

• A ∩ B = ∅,

• The number of elements of A is not an element of A,

• The number of elements of B is not an element of B.

Find N .


7. Jackie and Phil have two fair coins and a third coin that comes upheads with probability 4

7. Jackie flips the three coins, and then Phil

flips the three coins. Let mn

be the probability that Jackie gets thesame number of heads as Phil, where m and n are relatively primepositive integers. Find m+ n. (2010 AIME I Problem 4)

8. Define an ordered quadruple of integers (a, b, c, d) as interesting if1 ≤ a < b < c < d ≤ 10, and a+d¿b+c. How many ordered quadruplesare there? (2011 AIME I Problem 6)

9. Define an ordered triple (A,B,C) of sets to be minimally intersectingif |A ∩B| = |B ∩ C| = |C ∩ A| = 1 and A ∩B ∩ C = ∅. For example,({1, 2}, {2, 3}, {1, 3, 4}) is a minimally intersecting triple. Let N bethe number of minimally intersecting ordered triples of sets for whicheach set is a subset of {1, 2, 3, 4, 5, 6, 7}. Find the remainder when Nis divided by 1000.

Note: |S| represents the number of elements in the set S. (2010 AIMEI Problem 7)

10. LetN be the number of ways to write 2010 in the form 2010 = a3·103 =a2 ·102 = a1 ·10 = a0, where the ai’s are integers, and 0 ≤ ai ≤ 99. Anexample of such a representation is 1 ·103 = 3 ·102 = 67 ·101 = 40 ·100.Find N . (2010 AIME I Problem 10)

4

AIME Geometry Part I

Although synthetic geometry is, of course, extremely important for theAIME, it is nearly impossible to reliably approach geometry problems with-out a working knowledge of trigonometry. Even those who have no idea whatthis term means have surely heard of the “special” right triangles - in par-ticular those whose angles are 30-60-90 or 45-45-90. In a way, trigonometryis essentially an extension of these rules.

Trigonometry utilizes two basic functions - the sine function and thecosine function, abbreviated sin(x) and cos(x) respectively. The simplestdefinitions are

Definition 4.0.3. Construct a right triangle with an angle of θ. The sine ofθ, sin(θ), is the ratio of the side opposite θ to the hypotenuse of the triangle.The cosine of θ, cos θ, is the ratio of the side adjacent to θ to the hypotenuseof the triangle.

For example, the two aforementioned triangles gives us sin(30◦) = 12

and

cos(45◦) =√22

.This also gives us an obvious identity. The side opposite angle θ is also

adjacent to the angle 90− θ, and vice versa. As such we have

Theorem 4.0.15. For an angle θ, we have

sin(θ) = cos(90◦ − θ).

It is clear, however, that this definition quickly runs into serious issues.For example, we cannot represent sin(θ) as part of a right triangle when θis greater than 90◦. Instead we need to amend our definition slightly:

46


Definition 4.0.4. Consider the circle centered at the origin with radius 1.Let θ be the rotational distance from the positive x-axis to a line `. Thenthe line l intersects the circle at (cos θ, sin θ). Notice how this definition isessentially equivalen to the previous one, if we allow “negative lengths” (ordirected segments)

This allows us to extend our definition of sine and cosine to any angleθ, even those that are greater than 360◦. In that case, we are essentially“looping around” the circle. As such, we can relatively simply note that

Theorem 4.0.16. For any angle θ and any integer k, we have

sin(θ) = sin(360k + θ)

andcos(θ) = cos(360k + θ).

There are also several other relations between these functions. Noticethat if we reflect the line ` over the line y = x, the rotational distancebetween the positive x-axis and the image of the line ` becomes 90 − θ,meaning that our first theorem holds for all angles θ as well.

This reflection argument comes in handy for determining some otheridentities. For example, reflecting over the line x = 0 makes the rotationaldistance 180− θ and takes the point (cos θ, sin θ) to (− cos θ, sin θ). As such,we can conclude the following two identities:

Theorem 4.0.17. For any angle θ, we have

sin θ = sin(180− θ)

andcos θ = − cos θ.

Reflecting over the other axis, y = 0, sends θ to 360− θ and (cos θ, sin θ)to (cos θ,− sin θ). As such we can conclude (with the aid of our secondtheorem) that

Theorem 4.0.18.

cos(360− θ) = cos(θ) =⇒ cos(θ) = cos(−θ)


andsin(360− θ) = − sin(θ) =⇒ sin(−θ) = − sin(θ)

In other words, cosine is an even function and sine is an odd function.

These theorems allow us to put together a convenient table of commonangles:

θ (in degrees) sin(θ) cos(θ) θ sin(θ) cos(θ)0 0 1 180 0 -1

30 12

√32

210 −12

−√32

45√22

√22

225 −√22

−√22

60√32

12

240 −√32

−12

90 1 0 270 -1 0

120√32

−12

300 −√32

12

135√22

−√22

315 −√22

√22

150 12

−√32

330 −12

√32

One further identity, that is evident from our definitions of sine andcosine, is


sin2 θ + cos2 θ = 1

This was also evident from our earlier incomplete definition, as if we letthe sides be a, b, c with c being the hypotenuse, we would reach

sin2 θ + cos2 θ =(ac

)2+

(b

c

)2

=a2 + b2

c2= 1

by the Pythagorean Theorem.

In our table above, each of our angles are either multiples of 30◦ or 45◦

(or both). While convenient, the natural extension is whether we can findvalues for other angles. The answer, to some extent, is yes. Calculating thevalue of cos(1◦) is practically an impossible task, but perhaps something likesin(15◦) is reasonable. Indeed this is certainly doable; we need to utilize thesum and difference formulas for sine and cosine.


Theorem 4.0.20. For any angles α and β, we have

sin(α + β) = sinα cos β + sin β cosα

Because sin(−β) = − sin(β) and cos(−β) = cos(β), we also have

sin(α− β) = sin(α + (−β)) = sinα cos β − sin β cosα

For cosine, we have

cos(α + β) = cosα cos β − sinα sin β

andcos(α− β) = cosα cos β + sinα sin β.

Unfortunately, the proofs of these are not particularly motivated (thoughquite nice; see Wikipedia’s proof http://en.wikipedia.org/wiki/List_of_trigonometric_identities#Angle_sum_and_difference_identities).The “nicest” proof, per se, utilizes Euler’s identity

eiθ = cos θ + i sin θ

but this is too complex for this article (if you’re curious, multiply outei(α+β) = (eiα)(eiβ) and see what happens when you equate real/imaginaryparts). We can use this formula to routinely calculate

sin(15◦) = sin(45◦ − 30◦) = sin(45◦) cos(30◦)− sin(30◦) cos(45◦)

=

√2

2·√

3

2− 1

2·√

2

2=

√6−√

2

4

A special case of these identities are the double angle formulas:


sin(2θ) = 2 sin θ cos θ

andcos(2θ) = 2 cos2 θ − 1 = 1− 2 sin2 θ = cos2 θ − sin2 θ.

http://en.wikipedia.org/wiki/List_of_trigonometric_identities#Angle_sum_and_difference_identities

http://en.wikipedia.org/wiki/List_of_trigonometric_identities#Angle_sum_and_difference_identities


Notice that the notation cos2 θ means (cos θ)2, but cos−1 θ means some-thing very different (we’ll get to that later). If we solve for cos θ in the above,this gives us

cos θ = ±√

1 + cos(2θ)

2

which gives us

cos(θ

2) = ±

√1 + cos θ

2

where the ± is chosen based on the quadrant of θ. Plugging this back intoour sin2 θ + cos2 θ = 1 formula, we get

sin(θ

2) = ±

√1− cos θ

2

Naturally, these are termed the half angle formulas.This allows us, for example, to easily calculate sin(15◦). We have

sin(15◦) =

√1−

√32

2=

√2−√

3

4=

√6−√

2

4

exactly as we got before. Notice that we selected the positive solution assin(15◦) is clearly positive.

Before we move on, let’s collect these formulae into a handy referencesheet. In each of these, θ, α, β are any angles.

Basic Trigonometric Identities:

1. sin θ = cos(90− θ).

2. sin θ = sin(180− θ), while cos θ = − cos(180− θ).

3. sin θ = − sin(−θ), while cos θ = cos(−θ).

4. sin2 θ + cos2 θ = 1.

5. sin(α + β) = sinα cos β + sin β cosα and sin(α − β) = sinα cos β −sin β cosα.

6. cos(α + β) = cosα cos β − sinα sin β and cos(α − β) = cosα cos β +sinα sin β.


7. sin(2θ) = 2 sin θ cos θ, while cos(2θ) = 2 cos2 θ − 1 = 1 − 2 sin2 θ =cos2 θ − sin2 θ

8. sin(θ2

)= ±

√1−cos θ

2and cos

(θ2

)= ±

√1+cos θ

2, where we choose either

the positive or negative solution based on the quadrant of θ.

Understanding why these formulas are true is rewarding, but remem-bering their statements is of equal importance. If you understand themsufficiently you should be able to rederive them quickly, but having to dothis in the middle of a test costs precious minutes.

Only after examining these relationships between the sine and cosinefunctions does it make sense to introduction the tangent function, writtenas tan(θ). From our original definition, the tangent of an angle is the ratioof the side opposite to θ to the side adjacent to θ. Extending this to ourunit circle definition instantly gives us tan(x) = sin(x)

cos(x), exactly as expected

as ab

=acbc

. Most sources also define the reciprocal functions of the sine,

cosine, and tangent functions, terming the cosecant of θ to be the reciprocalof sine, so that csc(θ) = 1

sin(θ), the secant of θ to be the reciprocal of cosine,

so that sec(θ) = 1cos(θ)

, and the cotangent of θ to be the reciprocal of

tangent, so that cot(θ) = 1tan(θ)

. However, these are rarely useful beyondsimplifying a problem statement somewhat.

In general, the appearance of functions other than sine or cosine is asignal that you should rewrite everything in terms of sines and cosines assoon as it is computationally feasible. Nonetheless, the tangent functionhas some useful properties that are worth exploring on their own. Recallthat tan(θ) = sin(θ)

cos(θ), which immediately implies (from our list above) that

tan(θ) = − tan(180 − θ) = − tan(−θ). Furthermore, the relation sin2 θ +cos2 θ = 1 gives us the occasionally useful relations tan2 θ + 1 = sec2 θ and1+cot2 θ = csc2 θ (these follow by division by cos2 θ and sin2 θ respectively).

More interesting are the expressions for tan(α + β) and tan(α − β).Working this out is simply a matter of algebra, but it ends in a nice result.

tan(α + β) =sin(α + β)

cos(α + β)=

sinα cos β + sin β cosα

cosα cos β − sinα sin β.

Dividing through by cosα cos β gives

=

sinαcosα

+ sinβcosβ

1− sinα sinβcosα cosβ

=tanα + tan β

1− tanα tan β


Using the same strategy as before to handle differences of angles, we get

tan(α− β) = tan(α + (−β)) =tanα− tan β

1 + tanα tan β

The double angle formula for tangent is thus

tan(2θ) = tan(θ + θ) =2 tan θ

1− tan2 θ.

Finally, we conclude this article with the sum to product identities,which do exactly as you’d expect - take the sum of 2 trigonometric functionsand equate them to a product. For example, what could we do if we wantedto find an expression for sin(α+β)? The flash of inspiration here is to utilizethe average, which allows us to cancel some of the terms in the expansionof sin(α + β). Specifically,

sinα + sin β = sin

(α + β

2+α− β

2

)+ sin

(α + β

2− α− β

2

)

= sin

(α + β

2

)cos

(α− β

2

)+ sin

(α− β

2

)cos

(α + β

2

)+ sin

(α + β

2

)(α− β

2

)− sin

(α− β

2

)cos

(α + β

2

)= 2 sin

(α + β

2

)(α− β

2

)Repeating the same process backwards gives us the product-to-sum iden-

tity

sinα cos β =1

2[sin (α + β) + sin (α− β)]

Summary:At some point, every student became aware of the “special” 30-60-90

and 45-45-90 right triangles. But, in fact, there is nothing special aboutthese triangles at all! Trigonometry gives us a way to relate the ratios ofthe sides of any right triangle, regardless of its angles. A simple and natu-ral extension allows us to even drop the right triangle condition and simplywork with the two very important sine and cosine functions. As we saw,


these functions are highly related in several ways, and unsurprisingly theysatisfy several nice relations. Because trigonometry is ubiquitous in geom-etry, trying to get by on difficult AIME problems without at least havingtrig in your toolbox is close to suicide. In this, the first of two trigonometryarticles, we dealt with the “theoretical” side of things, focusing on formulaeand identities that will prove useful in a great deal of problems. The secondpart will focus more on their applications to problems, which as we will seeis not necessarily limited to geometry.

Exercises:

1. In the article we slightly skated over the fact that some of our formulas

(such as√

1−cos θ2

) may not be defined. Show that both sine and cosine

are functions from R→ [−1, 1], meaning that our relations are all valid(in other words, both sin(θ) and cos(θ) take any value of θ and returna value between −1 and 1 inclusive). What is the domain and rangeof the tangent function?

2. In the text, we saw that sin(2θ) = 2 sin θ cos θ and cos(2θ) = 2 cos2 θ−1. Using the same addition formulas, determine sin(3θ) and cos(3θ) interms of sin θ and cos θ.

3. We found half angle formulas for sine(√

1−cos θ2

)and cosine

(√1+cos θ

2

).

Find a half angle formula for tangent in terms of sine and cosine. Inother words, express tan

(θ2

)in terms of sin θ and cos θ.

4. At the end of the article, we transformed sinα + sin β into a product(and vice versa). Do the same for cosα + cos β.

5. Although we used degrees throughout this entire article for simplicity,trigonometry is best done in radians. Just as there are 360 degrees in acircle, we define a circle to contain 2π radians. This has the advantageof being a more natural definition, as the circumference of the unitcircle is exactly 2π. Convert the table containing the trigonmetricfunctions of angles that are multiples of 30◦ and/or 45◦ to radians.

6. We mentioned that although sin2 θ means (sin θ)2, sin−1 θ means some-thing entirely different. This represents the inverse sine, written aseither sin−1 θ or arcsin θ, which is the inverse of the sine function.For example, sin(π

6) = 1

2, so arcsin

(12

)= π

6. However, even though


sin(13π6

)= 1

2, we don’t have arcsin

(12

)= 13π

6. What is the domain

and range of arcsin? The function arccos is defined similarly, beingthe inverse of the cosine function. What is the domain and range ofthe arccos function? If the arctan function is defined as you’d expect,what is the domain and range of that function?

7. Determine the following compositions of trigonometric functions interms of the angle θ:

• sin[arccos(θ)]

• sin[arctan(θ)]

• cos[arcsin(θ)]

• cos[arctan(θ)]

• tan[arcsin(θ)]

• tan[arccos(θ)]

8. Determine a simple formula for the expression arctan(α) + arctan(β).

9. Let {an} be a sequence with |a1| < 1 and an = 2an−1√

1− a2n−1. Ifa6 = 1

2, find a12.

10. Find a simpler expression for cos θ cos(2θ) cos(4θ) . . . cos(2kθ).

5

AIME Geometry Part II

In the previous article, we saw several trigonometric identities. Before wemove on we’ll briefly summarize them below:

Basic Trignometric Identities:

1. sin θ = cos(90− θ).

2. (“Reflective” identities) sin θ = sin(180−θ), while cos θ = − cos(180−θ). Similarly, sin θ = − sin(−θ), while cos θ = cos(−θ).

3. (Pythagorean identities) sin2 θ + cos2 θ = 1, tan2 θ + 1 = sec2 θ, 1 +cot2 θ = csc2 θ.

4. (Sum and difference formulas; sine) sin(α+β) = sinα cos β+sin β cosαand sin(α− β) = sinα cos β − sin β cosα.

5. (Sum and difference formulas; cosine) cos(α + β) = cosα cos β −sinα sin β and cos(α− β) = cosα cos β + sinα sin β.

6. (Sum and difference formulas; tangent) tan(α + β) = tanα+tanβ1−tanα tanβ

and

tan(α− β) = tanα−tanβ1+tanα tanβ

.

7. (Double angle formulas) sin(2θ) = 2 sin θ cos θ, cos(2θ) = 2 cos2 θ−1 =1− 2 sin2 θ = cos2 θ − sin2 θ, tan(2θ) = 2 tan θ

1−tan2 θ .

8. (Half-angle formulas) sin(θ2

)= ±

√1−cos θ

2and cos

(θ2

)= ±

√1+cos θ

2,

where we choose either the positive or negative solution based on thequadrant of θ.

55


9. (Sum-to-product and Product-to-sum)

sinα + sin β = 2 sin

(α + β

2

)cos

(α− β

2

)and

sinα cos β =1

2(sin(α + β) + sin(α− β)).

The analogous expressions for other trigonometric functions were leftas an exercise in the previous article.

To these, we must add two very important identities about trianglesbefore we move on. The first is the Law of Sines:

Theorem 5.0.22 (Law of Sines). In a triangle ABC, we have

AB

sinC=

BC

sinA=

CA

sinB.

Using the standard notation AB = c, BC = a, and CA = b (e.g. a sidelength is represented by the lowercase version of the point it is opposite),we can write this more compactly as

a

sinA=

b

sinB=

c

sinC.

We can extend this to the (aptly named) Extended Law of Sines,which simply states that if R is the circumradius (the radius of the cir-cumcircle) of triangle ABC, we have

a

sinA=

b

sinB=

c

sinC= 2R.

This formalizes and extends some intuition we may have had about triangles,namely that the side lengths of a triangle are ordered in the same order astheir corresponding angles (so, for example, the longest side of a triangle isopposite its largest angle). Indeed, this is guaranteed by the Law of Sinesexactly. The second theorem is the Law of Cosines, an arguably even moreimportant theorem:


Theorem 5.0.23 (Law of Cosines). In a triangle ABC, we have

AB2 = AC2 +BC2 − 2(AC)(BC) cosACB.

Using standard lettering, we can write this in the nicer form

c2 = a2 + b2 − 2ab cosC.

Obviously, the analogous identities

b2 = a2 + c2 − 2ac cosB

a2 = b2 + c2 − 2ac cosA

hold as well.

We won’t provide a full proof of these theorems here, because they areeasily accessible and not particularly enlightening. Instead let’s convinceourselves of their truth by looking at the special case ∠ACB = 90◦ (e.g.angle C is right).

In this case, the Law of Sines gives us

a

sinA=

b

sinB=

c

sin 90◦= c

⇐⇒ sinA =a

cand sinB =

b

c

which is indeed true. Furthermore, when ABC is a right triangle with rightangle C, we also have R = c

2⇐⇒ 2R = c, as expected.

The Law of Cosines similarly gives us a well-known expression. We have

c2 = a2 + b2 − 2ab cos 90◦ = a2 + b2 − 2ab(0) = a2 + b2

which you should certainly recognize! In fact, in exactly this manner theLaw of Cosines tells us that c2 = a2+b2 if and only if ABC is a right trianglewith right angle C, meaning that we can think of the Law of Cosines as an“extended” Pythagorean theorem (which also leads to the natural proof).

After all this discussion, we are finally ready to dive into some prob-lems! Notice how all these AIME problems are in the late middle sectionof the test, but many of them are simply direct applications of trigonomet-ric manipulations. As such even a cursory knowledge of how to work with


trigonometric equations can easily score you some easy points. Also noticehow many of these problems are from different (but consecutive) years - inrecent years the AIME has usually had at least one “trigonometric manip-ulation” problem that just involves some algebra.

Problem 5.0.23. The domain of the function f(x) = arcsin(logm(nx)) isa closed interval of length 1

2013, where m and n are positive integers and

m > 1. Find the remainder when the smallest possible sum m+n is dividedby 1000. (2013 AIME I Problem 8)

Solution. This is hardly a difficult problem at all! Yet despite the fact thatit essentially tests your knowledge of some definitions, it is placed right inthe middle of the test. Anyway, let’s move on to the solution. Firstly weneed to start with the domain of arcsin(x). Even if you don’t remember itoffhand, it is very easy to rederive. We can easily recall that sin(x) takes allof the values from −1 to 1 inclusive, so arcsin(x)’s domain is [−1, 1] (and,to keep arcsin(x) a function, its range is [−π

2, π2]). Therefore we require

−1 ≤ logm(nx) ≤ 1

=⇒ 1

m≤ nx ≤ m

=⇒ 1

mn≤ x ≤ m

n

Now this becomes quite simple. We just need

m

n− 1

mn=

1

2013⇐⇒ m2 − 1

mn=

1

2013⇐⇒ 2013m2 − 2013 = mn

This implies that m is a factor of 2013. It is fairly obvious that we want tominimize m, so we choose m = 3 and thus n = 2013·8

3= 5368, meaning that

m+ n ≡ 3 + 5368 ≡ 371 (mod 1000).

All that was required in the above problem was to keep a level head andnot be deceived by the fancy notation. This transforms the problem intowhat is actually an elementary number theory problem, which is somethingwe’d hardly expect from a cursory glance at the problem statement. Wecontinue with some more “purely” trigonometric problems (e.g. those with-out direct geometric components) before seeing the value of the Law of Sinesand the Law of Cosines.


Problem 5.0.24. Let x and y be real numbers such that sinxsin y

= 3 andcosxcos y

= 12. The value of sin 2x

sin 2y+ cos 2x

cos 2ycan be expressed in the form p

q, where

p and q are relatively prime positive integers. Find p + q. (2012 AIME IIProblem 9)

Solution. This is another fairly straightforward problem, but it illustrates auseful general strategy. Oftentimes, when working with seemingly arbitraryequations involving sines and cosines, it is useful to square equations in orderto make use of the sin2 θ+ cos2 θ = 1 identity. This problem is no exception.We have

sinx = 3 sin y =⇒ 4 sin2 x = 36 sin2 y

2 cosx = cos y =⇒ 4 cos2 x = cos2 y

where we multiplied the first equation by 4 to make use of the cancellations.Adding these equations gives

4 = 1 + 35 sin2 y =⇒ sin2 y =3

35

This gives us sin2 x = 2735

, cos2 x = 835

, and cos2 y = 3235

. As such we have

sin 2x

sin 2y=

sinx cosx

sin y cos y=

√sin2 x cos2 x

sin2 y cos2 y=

√2735· 835

335· 3235

=

√9

4=

3

2

andcos 2x

cos 2y=

2 cos2 x− 1

2 cos2 y − 1=

2(

835

)− 1

2(3235

)− 1

=−19

352935

= −19

29

so we are looking for 32− 19

29= 87−38

2·29 = 4958

=⇒ 49 + 58 = 107 .

Problem 5.0.25. Suppose x is in the interval [0, π2] and log24 sinx(24 cosx) =

32. Find 24 cot2 x. (2011 AIME I Problem 9)

Solution. We’re only given one equation, so the only logical thing to do isto work with it. Let’s write

log24 sinx(24 cosx) =3

2=⇒ (24 cosx)2 = (24 sin x)3 =⇒ cos2 x = 24 sin3 x

Notice that we are looking for 24 cot2 x = 24 cos2 xsin2 x

= 576 sin x.


We can furthermore utilize the identity cos2 x = 1− sin2 x to get

24 sin3 x+ sin2 x− 1 = 0

At this point we can do a bit of meta-thinking. The question is asking for576 sinx, and the answer is necessarily an integer. This implies that sinx isrational, so we can search for rational roots of this equation. The RationalRoot Theorem tells us that we need to check ±1,±1

2,±1

3,±1

4,±1

6,± 1

12,± 1

24.

We quickly determine that sinx = 13

gives us a solution, implying that the

answer is 576 · 13

= 192 .For completeness, we can factor this cubic as

24 sin3 + sin2 x− 1 = (3 sinx− 1)(8 sin2 x+ 3 sinx+ 1) = 0

which has the only real solution sinx = 13.

Let’s take a step back from the math and look at the above two problemsagain. We’re talking about problems that test essentially just your basicknowledge of trigonometry and some algebraic manipulation, and these arein the mid-high range of the AIME. These are two problem 9s. Think aboutthat for a second. Scoring a 9 on the AIME means you have excellent chancesat qualifying for the USAMO (or the USAJMO) - all because you learnedsome basic trigonometry. Isn’t that amazing?

Problem 5.0.26. Find the positive integer n such that

arctan1

3+ arctan

1

4+ arctan

1

5+ arctan

1

n=π

4.

(2008 AIME I Problem 8)

Solution. This one requires some ingenuity. Unlike sine or cosine, we don’thave any tricky sum-to-product ideas for arctan. Using the average doesn’tseem to be helpful for anything either. All we really know about arctan isthat tan(arctan(θ)) = θ - but this is all we need!

Recall the addition formula for tangent:

tan(α + β) =tanα + tan β

1− tanα tan β

Now comes the tricky part. We can simplify this quite a bit by lettingα = arctanx and β = arctan y - this gives us

tan(arctanx+ arctan y) =tan(arctanx) + tan(arctan y)

1− tan(arctanx) tan(arctan y)=

x+ y

1− xy


taking the arctan of both sides gives us

arctanx+ arctan y = arctan(x+ y

1− xy).

This is exactly what we wanted! We can now calculate

arctan1

3+ arctan

1

4= arctan

( 13

+ 14

1− 13· 14

)= arctan

(7

11

)arctan

(7

11

)+ arctan

1

5= arctan

( 711

+ 15

1− 711· 15

)= arctan

(23

24

)Finally, we can note that tan π

4= 1 =⇒ π

4= arctan 1, so we need to solve

2324

+ 1n

1− 2324· 1n

= 1 =⇒ 23n+ 24

24n− 23= 1 =⇒ 23n+ 24 = 24n− 23 =⇒ n = 47 .

Problem 5.0.27. Let a = π2008

. Find the smallest positive integer n suchthat

2[cos(a) sin(a) + cos(4a) sin(2a) + cos(9a) sin(3a) + · · ·+ cos(n2a) sin(na)]

is an integer. (2008 AIME II Problem 8)

Solution. These look suspiciously like product-to-sum identities. In fact, wehave in general that

2 cos(n2a) sin(na) = sin(na+ n2a) + sin(na− n2a)

= sin (n(n+ 1)a)− sin (n(n− 1)a) .

But since n(n+ 1) = ((n+ 1)) ((n+ 1)− 1), we have a telescoping sum! Inother words,

n∑k=1

2 cos(k2a) sin(ka) =n∑k=1

[sin (n(n+ 1)a)− sin (n(n− 1)a)]

= sin(n(n+ 1)a)− sin(0) = sin (n(n+ 1)a)

which we need to be an integer. Notice that sin θ is an integer exactly whenπ2| θ, so we require π

2| n(n+1)π

2008, implying that n(n+1)

1004must be an integer.

Since n and n + 1 are relatively prime, it follows that exactly one ofthem is divisible by 251 (the largest prime dividing 1004). As such n ≡ −1(mod 251) or n ≡ 0 (mod 251). It is easy to check that n = 250 fails, whilen = 251 gives us n(n + 1) = 251 · 252 = 251 · 4 · 63, which is a multiple of251 · 4 = 1004.


Although such algebraic manipulations of trigonometric identities areubiquitous, the geometric side of things is at least as important. Rememberthe two formulas that we gave at the start of this article - they are extremelyimportant. Let’s start with a relatively simple example:

Problem 5.0.28. Equilateral triangle T is inscribed in circle A, which hasradius 10. Circle B with radius 3 is internally tangent to circle A at onevertex of T . Circles C and D, both with radius 2, are internally tangent tocircle A at the other two vertices of T . Circles B, C, and D are all externallytangent to circle E, which has radius m

n, where m and n are relatively prime

positive integers. Find m+ n. (2009 AIME II Problem 5)

Solution. Let the radius of circle E be r. We focus on triangle AED. Noticethat A,D, and the tangent point between circles A and D are collinear, sowe have AD = radius of A− radius of D = 10− 2 = 8. Similarly, ED goesthrough the point tangent to circles E and D, so we have ED = r + 2.Finally, EA = EB −AB = (r + 3)− (10− 3) = r − 4. Now we can use theLaw of Cosines, since we have ∠EAD = 60◦. This gives us

(r + 2)2 = (r − 4)2 + 82 − 2(r − 4)(8) cos 60◦

=⇒ r2 + 4r + 4 = r2 − 8r + 16 + 64− 8r + 32

=⇒ 20r = 108 =⇒ r =27

5giving the answer of 32.

The above example was not difficult - the hard part was setting up thetriangle to use the Law of Cosines on. To do so we need to be quite carefulin determining which points are collinear, as it is very possible to set up a“triangle” that doesn’t actually exist.

Now that we’ve seen the power of the Law of Cosines, let’s see an examplewith the Law of Sines. Notice how, at least in this author’s opinion, thebelow problem is an elementary exercise when using the tools in this articleand the previous one, yet it is placed very highly on the actual test. This isnot surprising - most high-numbered geometry questions on the AIME oftenhave relatively simple trigonometry solutions.

Problem 5.0.29. Let 4ABC be a right triangle with right angle at C. LetD and E be points on AB with D between A and E such that CD and CEtrisect ∠C. If DE

BE= 8

15, then tanB can be written as

m√p

n, where m and n

are relatively prime positive integers, and p is a positive integer not divisibleby the square of any prime. Find m+ n+ p. (2012 AIME I Problem 12)


Solution. The first thing we need to do is dissect the language. Trisectionsounds scary at first glance, but since C is a right angle all it’s saying is that∠ACD = ∠DCE = ∠ECB = 30◦.

We’re also given that DEBE

= 815

. Since DE and BE are both opposite30◦ angles, this problem screams for the Law of Sines. Of course, these sidesare not in the same triangle, so we need two applications of this theorem.Applying the Law of Sines to triangle CDE gives us

DE

sin 30◦=

EC

sin∠CDE

(=

DC

sin∠DEC

)and to BCE gives us

BE

sin 30◦=

EC

sin∠CBE

(=

BC

sin∠BEC

)Our next step is no surprise - the first equality in these chains is clearly themost helpful. We get

EC sin 30◦ = DE sin∠CDE = BE sin∠CBE

=⇒ DE

BE=

sin∠CBEsin∠CDE

We’re already given that DEBE

= 815

, so this simplifies to

8 sin∠CDE = 15 sin∠CBE

Now we need to take a second to assign some variables for convenience. Sincewe are specifically asked for tanB = tan∠CBE, it makes sense to denote∠CBE = θ. This gives us ∠CDE = ∠CDB = 180◦ − ∠DCB − ∠CBD =180◦ − 60◦ − θ = 120◦ − θ. Thus we can rewrite our equation as

8 sin(120◦ − θ) = 15 sin θ

=⇒ 8(sin 120◦ cos θ − sin θ cos 120◦) = 15 sin θ

=⇒ 8

(cos θ√

3

2− sin θ

2

)= 15 sin θ

=⇒ 4 cos θ√

3 = 11 sin θ

=⇒ tan θ =4√

3

11

making the answer 18 .


In the first problem, we needed just a single application of the Law ofCosines. In the above one, we needed two applications of the Law of Sines.To conclude this article, we look at a problem where three applicationsof the Law of Cosines is necessary to finish the problem. However, theseapplications are very motivated, and so such a solution is actually veryshort in a contest environment.

Problem 5.0.30. A paper equilateral triangle ABC has side length 12.The paper triangle is folded so that vertex A touches a point on side BCa distance 9 from point B. The length of the line segment along which thetriangle is folded can be written as

m√p

n, where m, n, and p are positive

integers, m and n are relatively prime, and p is not divisible by the squareof any prime. Find m+ n+ p. (2013 AIME I Problem 9)

Solution. Let’s first assign some variables for convenience. Let the pointthat A folds to be D, and the crease be EF with E on AC and F onAB. Let AE = x,AF = y, EF = z. This immediately implies that CE =12− x,BF = 12− y. Furthermore DF = AF = x and DE = AF = y.

It is fairly obvious that we should use the Law of Cosines on trianglesBDF and CED, as this will allow us to completely determine x and y. Let’sstart with triangle BDF . Since ∠FBD = 60◦, we get

x2 = (12− x)2 + 92 − 2(9)(12− x) cos(60◦)

=⇒ x2 = x2 − 24x+ 144 + 81− 108 + 9x

=⇒ 15x = 117 =⇒ x =39

5.

Doing a similar process for 4ECD gives us

y2 = (12− y)2 + 32 − 2(3)(12− y) cos(60◦)

=⇒ y2 = y2 − 24y + 144 + 9− 36 + 3y

=⇒ 21y = 117 =⇒ y =39

7.

Now we can finish using the Law of Cosines on 4DEF , since ∠EDF =60◦. We get

EF 2 =

(39

5

)2

+

(39

7

)2

− 2

(39

5

)(39

7

)cos 60◦


=⇒ EF 2 =392 · 72 + 392 · 52 − 392 · 5 · 7

52 · 72

=⇒ EF =39√

39

35

making our answer 39 + 39 + 35 = 113.

Summary:In the first Geometry article, we introduced trigonometry and discussed

various identities and the like. In this article, we got down to applying ournewfound knowledge to AIME problems. We first went through the alge-braic portion, looking at problems ranging from pure manipulations to asimplification to a telescoping sum utilizing the sum-to-product (and viceversa) identities. We then moved on to the geometric problems, which useeither (or both!) the Law of Sines or the Law of Cosines, often multipletimes. As is evident from the vast number of AIME problems in this arti-cle, the AIME extensively tests trigonometry (sometimes unintentionally!),including many problems in the 11-15 range. But, as we saw, even a passingknowledge of trigonometric manipulations (and in some cases simply defini-tions!) was enough to successfully tackle several mid-level AIME questions.

Exercises:

1. Point D lies on side AC of equilateral triangle ABC such that themeasure of angle DBC is 45 degrees. What is the ratio of the area oftriangle ADB to the area of triagle CDB? Express your answer as acommon fraction in simplest radical form. (MATHCOUNTS)

2. Determine the value of cos 1◦ + cos 2◦ + . . .+ cos 358◦ + cos 359◦

3. In triangle ABC, side AC and the perpendicular bisector of BC meetin point D, and BD bisects angle ABC. If AD=9 and DC=7, whatis the area of triangle ABD? Express your answer in simplest radicalform.

4. In rectangle ABCD, AD is twice AB, N is the midpoint of AD, andM is the midpoint of BN . What is the value of ∠NMD + ∠NBD?

5. Find the least positive integer n such that

1

sin 45◦ sin 46◦+

1

sin 47◦ sin 48◦+ · · ·+ 1

sin 133◦ sin 134◦=

1

sinn◦.


6. Kelvin the frog is building himself a new home. He has exactly 12units of fencing, but because frogs are not particularly intelligent hecan only construct polygons. Furthermore, each side must have inte-ger length. What is the maximum area that Kelvin the frog can make?

(A) 4√

3 (B) 9 (C) 6√

3 (D) 6 + 3√

3 (E) 36π

7. What is the sum of all positive real solutions x to the equation

2 cos(2x)

(cos(2x)− cos

(2014π2

x

))= cos(4x)− 1?

(A) π (B) 810π (C) 1008π (D) 1080π (E) 1800π(2014 AMC 12B Problem 25 (!))

8. Let A1A2A3A4A5A6A7A8 be a regular octagon with side length 1, and

P be a point in its interior. The maximum possible value of8∑i=1

AiP2

can be written as m + n√p, where m and n are positive integers and

p has no perfect square factors other than 1. What is m+ n+ p?

(A) 10 (B) 14 (C) 18 (D) 22 (E) 26

9. Let {an}Nn=1 be a finite sequence of increasing positive real numberswith a1 < 1 such that

an+1 = an

√1− a21 + a1

√1− a2n

and a10 = 12. What is a20? (2011 Duke Math Meet)

10. Let a0, a1, a2, . . . be an infinite sequence of real numbers such thata0 = 4

5and

an = 2a2n−1 − 1

for every positive integer n. Let c be the smallest number such thatfor every positive integer n, the product of the first n terms satisfiesthe inequality

a0a1 . . . an−1 ≤c

2n.

What is the value of 100c, rounded to the nearest integer? (2013 MathPrize for Girls Problem 20)

6

AIME Number Theory Part I

How many times have we all seen problems along the lines of “find the lastdigit of (some exponential)”, such as “find the last digit of 32004?” If you’veever participated in MATHCOUNTS or similar competitions, the answer isa lot. The usual strategy on this problem is to examine the powers of 3modulo 10:

n 3n 3n (mod 10)1 3 32 9 93 27 74 81 15 243 36 729 97 2187 78 6561 1

This reveals a simple pattern - the last digits of the powers of 3 follow thecycle 3, 9, 7, 1, 3, 9, 7, 1, . . .. Since 2004 ≡ 0 (mod 4), we know that 32004 ≡ 34

(mod 10), making our answer 1.This is well and good, but what if we wanted something like the last two

digits of 72004? Let’s try our listing again:

n 7n 7n (mod 100)1 7 72 49 493 343 434 2401 15 16807 7

67


As soon as we reach 74 ≡ 1 (mod 100), we can stop, because this imme-diately tells us that 7n+4 ≡ 7n (mod 100). So, once again, we have 72004 ≡ 1(mod 100). But the case of 7 is rather special - what would happen if, forexample, we tried to find the last two digits of 22014?

n 2n 2n (mod 100) n 2n 2n (mod 100)1 2 2 12 4096 962 4 4 13 8192 923 8 8 14 16384 844 16 16 15 32768 685 32 32 16 65536 366 64 64 17 131072 727 128 28 18 272144 448 256 56 19 524288 889 512 12 20 1048576 7610 1024 24 21 2097152 5211 2048 48 22 4194304 4

Finally! We get 222 ≡ 22 (mod 100). This implies that, when n ≥ 2, wehave 2n+20 ≡ 2n (mod 100). This isn’t 100% satisfying, because unlike theother powers this cycle does not contain a 1 (which would immediately implya cycle). But this is obviously not going to happen, as 2n is always even andthus cannot be 1 (mod 100). While, again, this is all well and good, thisis a rather unenlighening approach. If I wanted, for example, to now knowthe last 2 digits of 32014, repeating this process does not exactly sound likea fun or efficient task. Instead we should look for a more general tool.

As usual when facing such a task, we attempt to formally state theproblem. First, it is useful to denote the following definition:

Definition 6.0.5. For relatively prime integers a and n, the order of amodulo n, denoted ordna, is the smallest positive integer k such that ak ≡ 1(mod n).

For example, ord103 = 4, since 34 ≡ 1 (mod 10) and none of 31, 32, 33

are 1 (mod 10). Now we need to determine precisely what it is we wantto determine about this definition. Obviously, the optimal situation is thatwe find a specific and simple formula for ordna. But even from just whatwe’ve found above, this seems highly unlikely. For example, ord252 = 20,but ord257 = 4. To make things even more challenging, we can find easythings such as ordn(n− 1) = 2. All this means that there is likely no simpleformula.


Next up on our wishlist is to find a formula for some k, not necessar-ily the smallest, such that ak ≡ 1 (mod n). Fortunately, this is certainlydoable, but it requires some creativity. Let’s look at a seemingly completelyunrelated problem:

Problem 6.0.31. Let Sn be the set {1, 2, 3, . . . , n− 1}, and aSn be the set{a, 2a, . . . , a(n− 1)}, where each of the elements are reduced modulo n. Forwhich a1, a2, n is a1n and a2n the same set?

This notation may seem extremely scary (especially since I arbitrarilymade it up - the notation Fp is far more standard), so let’s look at a briefexample. The set S5 is {0, 1, 2, 3, 4} - nothing scary about that. If we take2S5, or the set {0, 2, 4, 6, 8}, and reduce each of the terms modulo 5, theset becomes {0, 2, 4, 1, 3}. This is, up to order, the same as S5, so we cansay S5 = 2S5. This is, of course, not always the case. Moving up a numberto S6 = {0, 1, 2, 3, 4, 5}, we get 2S6 = {0, 2, 4, 6, 8, 10} =⇒ {0, 2, 4, 0, 2, 4},which is not the same as S6. However, 5S6 = {0, 5, 10, 15, 20, 25} =⇒{0, 5, 4, 3, 2, 1} is indeed S5.

Before reading on, play around with some of these sets (take S = 3, 4, 5, 6, 7as a start) and make some observations. See how many of the below younoticed, even if they’re vaguely represented in your mind.

• No matter what we multiply by, we seem to get something alongthe lines of a cycle. For example, even though 2S6 6= S6, the set{0, 2, 4, 0, 2, 4} does reduce to {0, 2, 1, 0, 2, 1} upon taking a modulusof three, making this, in some sense, two “copies” of S3. Informally, ifwe made a square table of values, it’s “nice”. Don’t worry if all thisseems rather informal and quite stupid; capturing a thought processin words is very difficult.

• Regardless of our choice for n, we always have Sn = (n− 1)Sn.

• When n is prime, it appears aSn = Sn for any choice of a, unless ahappens to be a multiple of p.

• More generally, when a and n are relatively prime it appears aSn = Sn.

The latter two observations are key. Let’s prove them rigorously.

Problem 6.0.32. L Sp denote the set {1, 2, . . . , p− 1}, where p is a primenumber. Let aSp denote the set {a, 2a, . . . , a(p−1)} after taking each of theelements modulo p. Show that if p - a, then aSp = Sp.


Solution. To do this, we can utilize a clever argument. Notice that the set{a, 2a, . . . , a(p− 1)} contains exactly p− 1 elements, and none of them aremultiples of p. Thus, if we can show that none of them are equal modulop, then this implies the result (why?). Fortunately, this is obvious. If wehave k1a ≡ k2a (mod p) for some 0 ≤ k1, k2 ≤ p− 1, then we can remembermodular division to get k1 ≡ k2 (mod p

gcd(p,a)). But because p - a we have

gcd(p, a) = 1, meaning that k1 ≡ k2 (mod p). However, 0 ≤ k1, k2 ≤ p− 1,so this is possible only if k1 = k2.

As such, each of the numbers a, 2a, . . . , a(p − 1) represent a differentresidue modulo p, and so aSp = Sp as desired.

So how on earth does this relate to our original problem at all? Theanswer is simple - it (almost) immediately solves it! How can this be? Well,let’s let PS be the product of the elements in S. Since aSp = Sp, we obviouslyhave PaSp = PSp . In other words,

1 · 2 · 3 · . . . · (p− 1) ≡ a · 2a · 3a · . . . · a(p− 1) (mod p)

=⇒ 1 · 2 · 3 · . . . · (p− 1) ≡ ap−1 · 1 · 2 · . . . · (p− 1) (mod p)

Notice that gcd(p, 1) = gcd(p, 2) = . . . = gcd(p, p − 1) = 1, so we candivide through by (p− 1)! to get

Theorem 6.0.24. For a prime number p and a number a relativelyprime to p, we have ap−1 ≡ 1 (mod p).

This is sometimes written as ap ≡ a (mod p), because this allows usto drop the “relatively prime” condition. Should a and p not be relativelyprime, this directly implies that p | a, implying that ap ≡ a ≡ 0 (mod p).

This is known as Fermat’s Little Theorem, or FLT (not to be con-fused with Fermat’s Last Theorem, which is slightly more difficult to demon-strate!). Despite its apparent simplicity, it is shockingly applicable in a va-riety of problems. At first glance, the fact that it is restricted to prime pmakes its usage seem rather contained, but utilizing some basic modulararithmetic laws expands it power dramatically. For example, consider theexample we gave at the beginning of this chapter (compute 32004 (mod 10)).We saw that the powers of three cycle every four terms. A more remarkableresult is that this holds true in general! Specifically,

Problem 6.0.33. For given numbers a, n, what is the last digit of an?


Solution. We are looking for an (mod 10). At first glance, we don’t yet haveany useful tools to approach this problem, because 10 is not prime. But wecan employ a very common technique here - split the modulus into primes.In other words, let’s look at an (mod 2) and an (mod 5).

FLT tells us that a1 ≡ 1 (mod 2) when a is relatively prime to 2 - let’swrite this as a2 ≡ a (mod 2) to avoid confusion (FLT does not claim thatevery integer is odd. It does, however, claim that every integer relativelyprime to 2 is odd, which is obviously true). It also tells us that a4 ≡ 1(mod 5) when a is relatively prime to 5. Again, let’s write this as a5 ≡ a(mod 5) to avoid having to deal with relatively prime shenanigans.

Combining these immediately tells us that a4n+k ≡ ak (mod 2) anda4n+k ≡ ak (mod 5). As such, we can conclude that a4n+k ≡ ak (mod 10),which is just a fancier way of saying that the last digits of all powers repeatevery 4 terms. Notice how we sometimes need to specify “for sufficientlylarge k” - we cannot, for example, say that 24 ≡ 20 (mod 10). This isbecause gcd(2, 10) 6= 1 - saying 34 ≡ 1 (mod 10) would indeed be a validstatement. But taking the time to ensure we don’t miss small cases is wellworth the effort, as we shall see in a certain upcoming AIME problem.

With our newfound tools, surely we can obliterate the next problem welooked at, namely the extension to the last two digits! Let’s split up 100 inprime factors - 100 = 22 · 52. Erm.

Unfortunately, FLT doesn’t really work as well here. Sure, we can saya4 ≡ 1 (mod 5), but there’s no convenient way to split 25 into moduli. Weare, once again, stuck. Back to our observations about random sets! Recallthat it appeared that aSn = Sn when a, n are relatively prime. If we couldprove that, maybe we could use an analogous method to the proof of FLT...

Problem 6.0.34. Let n be an integer, and Sn denote the set {1, 2, 3, . . . , n−1} (again, we ignore 0). Let aSn denote the set {a, 2a, . . . , a(n − 1)} aftertaking each of the elements modulo n. Show that if a, n are relatively prime,then aSn = Sn.

Solution. Just like before, if we could prove that a, 2a, . . . , a(n − 1) are alldistinct modulo n, then we’d be immediately done. Let’s assume otherwise,e.g. that there exist k1, k2 such that k1a ≡ k2a (mod n). This implies thatk1 ≡ k2 (mod n), since gcd(a, n) = 1. So now we’re done, right? By thesame logic as before, we get an−1 ≡ 1 (mod n) for any a relatively prime ton.

Well, no. As a direct counterexample, 34−1 ≡ 3 (mod 4), not 1. Whereis our error? Well, in the proof of FLT we divided through by (p−1)!, which


was valid as (p − 1)! and p are relatively prime. But this is not generallytrue of (n − 1)! and n (in fact it’s only true when n is prime or 1), so weneed to rethink our approach.

As we saw above, the fact that PSp and p were relatively prime was akey part of our solution. So perhaps we could define a new type of set - onewhere this is guaranteed?

Problem 6.0.35. Let Tn denote the set of all integers less than n that arerelatively prime to n. Let aTn denote the set formed by multiplying each ofTn’s elements by a, then taking their remainder modulo n. Show that if ais relatively prime to n, then Tn = aTn.

Before we move on the solution, it is useful to introduce the followingdefinition:

Definition 6.0.6. For any integer n, define φ(n) to be the number of integersless than or equal to n that are relatively prime to n. We say “or equal to”in order to establish φ(1) = 1.

This is known as Euler’s totient function, or often Euler’s phi func-tion.

In other words, φ(n) is the size of Tn.

Solution. Let the elements of Tn be t1, t2, . . . , tφ(n). Since these are all rel-atively prime to n, and a is also relatively prime to n, all the numbersat1, at2, . . . , atφ(n) are also relatively prime to n. This implies that all theelements of aTn are relatively prime to n, so it suffices to show that no twoof them are equal modulo n.

Now we can use the same method as above. If some k1, k2 existed withk1a ≡ k2a (mod n), then we could divide through by a to get k1 ≡ k2(mod n), a contradiction. Since there are φ(n) elements in each of the sets,we get

PTn ≡ PaTn = aφ(n)PTn (mod n)

and since all the elements of Tn are relatively prime to n, we can dividethrough by PTn to get

Theorem 6.0.25. For any integer a relatively prime to n, we have


aφ(n) ≡ 1 (mod n).This beautiful equation is known as Euler’s Theorem (one of many!).

Notice that all the numbers 1, 2, 3, . . . , p− 1 are relatively prime to p, soφ(p) = p−1. As such, ap−1 ≡ 1 (mod p), which is exactly FLT. This meansthat Euler’s Theorem is actually a generalization of FLT.

Let’s look at some applications of this. Recall that all of this nonsensewas started by our insistence on a general method for approaching the fol-lowing problem:

Problem 6.0.36. Determine the last 2 digits of 22014.

Solution. By now, we must have the tools to obliterate this problem withoutan ounce of effort. Notice that 22014 ≡ 0 (mod 4), so we just need to find22014 (mod 25).

Since 2 and 25 are relatively prime, we can apply Euler’s theorem to get2φ(25) ≡ 1 (mod 25). And now we... we... check every single number lessthan 25 to see if it’s relatively prime to 25?

After inserting the obligatory “curses, foiled again!”, we go back to thedrawing board. The relation aφ(n) ≡ 1 (mod n) is beautiful and all, but it’snot practically useful unless we have an easy way of determining φ(n). Let’stry and figure this out! As usual, we start with simple cases. If p is a primenumber, then clearly φ(p) = p− 1 (we already stated this above). Similarly,if n = pk for some prime p and exponent k, then φ(n) = pk − pk−1 =

pk−1(p− 1) = n(

1− 1p

).

Let’s take a look at what happens when n is the product of two primes,namely pq. Then there are pq total numbers between 1 and pq, q of whichare divisible by p, p of which are divisible by q, and 1 of which is divisibleby both. By PIE, there are thus pq − p − q + 1 = (p − 1)(q − 1) numbersrelatively prime to n, meaning φ(pq) = (p− 1)(q − 1). By similar logic, we

can show φ(pk1qk2) = pk1qk2(

1− 1p

)(1− 1

q

).

We might theorize at this point that

φ(pe11 pe22 . . . pekk ) = pe11 p

e22 . . . pekk

(1− 1

p1

)(1− 1

p2

). . .

(1− 1

pk

)and we’d be correct! But how can we rigorously prove this?


Problem 6.0.37. Prove the formula for φ(n) above. In particular, if the

prime factors of n are p1, p2, . . . , pn, show that φ(n) = n(

1− 1p1

)(1− 1

p2

). . .(

1− 1pk

).

Solution. Because we already showed φ(pk) = n(

1− 1p

), it would suffice

to show that φ(nm) = φ(n)φ(m) when n,m are relatively prime. This isknown as the phi function being multiplicative.

Consider the sets Tn and Tm, where T is defined as above in the article.By the Chinese Remainder Theorem, which sounds scary but is actually arelatively simple statement, for every x ≡ a (mod n) and x ≡ b (mod m)with m,n relatively prime, there is a unique c such that x ≡ c (mod mn).What this says is that, if |S| denotes the number of elements in S, |Tn||Tm| =|Tnm|. But by definition, there are φ(n), φ(m), and φ(mn) such numbersrespectively, so φ(m)φ(n) = φ(mn) as desired.

Before we finish up, we’ll briefly state the Chinese Remainder Theorem(without proof), as it’s a useful thing to know on some problems:

Theorem 6.0.26. Suppose n1, n2, . . . , nk are pairwise relatively rpime.In other words, there do not exist i, j such that gcd(ni, nj) 6= 1. Then,for any sequence of integers a1, a2, . . . , ak, there exists an integer x suchthat x ≡ ai (mod ni) for all 1 ≤ i ≤ k. Furthermore, all such solutionsare equivalent modulo n1n2 . . . nk.

In particular, if we have a set of congruences n ≡ a1 (mod b1), n ≡a2 (mod b2), . . . , n ≡ ak (mod bk), where b1, b2, . . . , bk are pairwise rela-tively prime, then there exists exactly one x such that n ≡ x (mod b1b2 . . . bk)and 0 ≤ x < b1b2 . . . bk.

NOW, we MUST be able to finish our original problem.

Problem 6.0.38. Determine the last 2 digits of 22014.

Solution. As before, we have 22014 ≡ 0 (mod 4). It remains to calculate 22014

(mod 25). Euler’s formula tells us that 2φ(25) ≡ 1 (mod 25), but now we canfinally determine precisely what φ(25) is! We have φ(25) = 25

(1− 1

5

)= 20,

so 220 ≡ 1 (mod 25). This implies 22014 ≡ 214 (mod 25), which we canquickly determine (by rote calculation or otherwise) to be 9 (mod 25).

Thus, 22014 ≡ 0 (mod 4) and 22014 ≡ 9 (mod 25). Putting these to-gether, we get 22014 ≡ 84 (mod 100). Notice that this is the exact sameanswer as we got from bashing out a pattern above.


As a final note, why did we split 100 into its prime factors (or powers ofprime factors, as it were)? Why did we not directly compute φ(100)? Ofcourse, we could have done precisely this - since φ(100) = 40, the statementa40 ≡ 1 (mod 100) (where a is relatively prime to 100) is indeed absolutelytrue. However, by splitting 100 into its prime powers, we end up with thestronger statement a20 ≡ 1 (mod 100). Why do we get a stronger bound?

This, in a sense, is how real mathematics is done. Recall that all of thiscomplicated stuff, with scary names like “Euler’s phi function” and “Chineseremainder theorem”, all came about because of our insistence on an “eas-ier” way of computing 22014 (mod 100). Hopefully, the material was laid outin a way that makes it all seem, if not natural, at least highly motivated.I also intentionally avoided accepted notation throughout this exploration.Although this may, at times, have made the explanations a little harderto follow, it emphasizes the process rather than the end result. Keep thisin mind the next time you work on a seemingly simple problem, whetherfrom MATHCOUNTS or the USAMO. Generalizations are almost alwayspossible, and usually far more interesting. Play around a bit, bash out somenumbers in the hopes of finding a pattern (often with the help of some pro-gramming), and you’ll hopefully come up with something cool.

Summary:

The infamous “last digit” problem disguises a whole new area of math-ematics. Although most students will, at least in passing, be familiar withthe concept that powers “cycle in groups of 4”, few understand why thisholds, and fewer still know how to extend this process. Starting with such asimple question like “what are the last two digits of 22014?” naturally leadsto a whole new set of questions, each of which open up more paths of explo-ration in their solutions. The most important results are Fermat’s LittleTheorem, stating that ap−1 ≡ 1 (mod p) for any prime p and a relativelyprime to p, and its generalization Euler’s Theorem, which takes this re-lation to the next level, stating aφ(n) ≡ 1 (mod n) when a, n are relativelyprime. As we may see in a later article, even these are just starting points.Although these require a great deal of creativity to prove, even the simpleconcept of focusing on prime factors is an incredibly important one.

Exercises:

1. Notice that the last digit of the powers of a repeat every four terms,the last 2 digits repeat every 20 terms, and the last 3 digits repeat


every 100 terms. Guess a general formula for how many terms it takesthe last k digits of the powers of a to repeat (for every a), and proveit.

2. A number a is said to be a primitive root modulo n if the numbersa1, a2, a3, . . . take on every possible residue modulo n (where “possible”means any integer less than n and relatively prime to n). Show thatif a is a primitive root, then orda(n) = φ(n).

3. Given any a, does there always exist a b such that ab ≡ 1 (mod 5)?What about ab ≡ 1 (mod 6)?

4. For what n does it appear to be true that for any a there exists a bsuch that ab ≡ 1 (mod n)?

5. We say two polynomials are equivalent modulo p if every correspondingcoefficient is equivalent modulo p. Obviously this definition only makessense when we are talking about polynomials with integer coefficientsthat accept only integers as input. For example, the polynomials x2 +2x+ 1 and x2 + 1 are equivalent modulo 2, but x3 + 2x2 + 2x+ 1 and1 are not. Suppose that p(x) and q(x) are equivalent modulo n. Showthat if r is a root of p(x), then r is also a root of q(x), and vice versa.Notice that, working over modulo n, if p(x) has a root r and q(x) hasa root r + n, the polynomials are still said to share a root.

6. Show that (a+ b)p ≡ ap + bp when working over modulo p.

7. What is the maximum number of roots a polynomial can have whenworking over modulo n? What polynomial has these roots?

8. Working over modulo p, where p is a prime, what roots does P (x) =xp−1 − 1 have?

9. Let the roots of xp−1 − 1 be r1, r2, . . . , rk. Let Q(x) = (x − r1)(x −r2) . . . (x − rk). What is the degree of the polynomial P (x) − Q(x)?How many roots does it have? What does this tell us?

10. Show that (p − 1)! ≡ −1 (mod p) for any prime p. This is known asWilson’s Theorem.

7

AIME Number Theory Part II

In this article, we work through many of the Number Theory problemspresent in recent AIME years. Most of these are of the form “set up the rel-evant modular equations and solve”, but knowing the tools and shortcuts todo this is necessary in a contest environment. Unlike other articles, insteadof first talking about results and then applying them, we will dive right intoproblems and focus on general results if and when we need them. Let’s startwith a classic example:

Problem 7.0.39. How many of the integers between 1 and 1000, inclusive,can be expressed as the difference of the squares of two nonnegative integers?(1997 AIME I Problem 1)

Solution. We want to know which numbers can be expressed in the formx2 − y2. As usual when working with squares, especially two squares, wework over modulo 4.

We can easily verify that x2 ≡ 0 or 1 (mod 4), and similarly y2 ≡ 0, 1(mod 4). Thus x2− y2 = 0,−1, or 1 (mod 4). In particular, x2− y2 cannotever be 2 (mod 4).

As such exactly 250 numbers cannot be expressed in this form, so ouranswer is 750 . For completeness, every odd integer 2k−1 can be expressedin the form k2 − (k − 1)2 and every multiple of 4, 4k, can be expressed inthe form (k + 1)2 − (k − 1)2.

Don’t be fooled by the fact that this was a question 1. It’s a more difficultquestion than many mid-level AIME questions today, but still requires nomore than some playing and a little intuition. Examining mod 4 becomesvery natural when examining some small cases, and indeed this immediatelyfinishes the problem.

77


Problem 7.0.40. Maya lists all the positive divisors of 20102. She thenrandomly selects two distinct divisors from this list. Let p be the probabilitythat exactly one of the selected divisors is a perfect square. The probability pcan be expressed in the form m

n, where m and n are relatively prime positive

integers. Find m+ n. (2010 AIME I Problem 1)

Solution. This problem illustrates the importance of working with primenumbers. Let’s do precisely this - 20102 = (2 · 3 · 5 · 67)2 = 22 · 32 · 52 · 672.How many divisors does this number have? Every divisor is of the form2a13a25a367a4 , where a1 ≤ 2, a2 ≤ 2, a3 ≤ 2, a4 ≤ 2. Since of each of thesenumbers can be 0, 1, or 2, there are a total of 34 = 81 divisors.

Of these, which are perfect squares? A perfect square means that eachof these exponents is even, so there are two choices (namely 0 or 2) for eachexponent. Thus there are a total of 24 = 16 perfect square divisors.

We can now directly compute the probability as (the first case is theprobability of picking a perfect square factor first and a non-perfect squarefactor after, while the second is the other way around) 16

81· 6580

+ 6581· 1680

=2 · 16

81· 6580

= 2681

, making our answer 26 + 81 = 107.

The above utilizes the following fact:

Theorem 7.0.27. Let the prime factorization of N be pe11 pe22 . . . penn . If

N is a perfect kth power, then k | e1, k | e2, . . . , k | en.

For example, if we wanted to find the smallest number n such that n · 8!were a perfect cube, we would prime factorize this as n·23 ·7·(2·3)·5·22 ·3·2 =27 · 32 · 5 · 7 ·n, meaning that the smallest possible n is 22 · 3 · 52 · 72 = 14700.

Problem 7.0.41. There are nonzero integers a, b, r, and s such that thecomplex number r+si is a zero of the polynomial P (x) = x3−ax2 +bx−65.For each possible combination of a and b, let pa,b be the sum of the zeroesof P (x). Find the sum of the pa,b’s for all possible combinations of a and b.(2013 AIME I Problem 10)

Solution. If you don’t understand some of the statements in this solution,you might want to read the Algebra article first! Since r+ si is a root of thepolynomial, so is r − si. The polynomial is a cubic, so there must be somereal root c. By Vieta’s formulas we have

c(r + si)(r − si) = c(r2 + s2) = 65


Notice that 65 factors into 5 · 13, so we have four cases. Before diving intothem, let’s first establish some other results:

• The solutions will be of the form (c,±r,±s). However, we only fo-cus on the positive solution for s, because the solutions (c,±r, s) and(c,±r,−s) represent the exact same polynomial - the one with rootsc,±r + is, and ±r − is.

• The sum of the roots is c+ (r + si) + (r − si) = c+ 2r.

• Because r can be positive or negative, the r part of this cancels out.As such, for any case where c = k and r2 + s2 = 65

k, the total sum

contributed is k · (number of solutions to r2 + s2 = 65k

).

Now we can simply do casework:

• Case 1: c = 1, r2 + s2 = 65. Here there are exactly eight solutions, as(r, s) = (±8, 1), (±7, 4), (±4, 7), or (±1, 8). Thus this case contributes8 to the sum of the pa,bs.

• Case 2: c = 5, r2 + s2 = 13. Here there are four solutions, as (r, s) =(±3, 2) or (±2, 3). Thus this case contributes 20 to the sum.

• Case 3: c = 13, r2 + s2 = 5. Here there are again four solutions, as(r, s) = (±2, 1) or (±1, 2). This case thus contributes 52 to the sum.

• Case 4: c = 65, r2 + s2 = 1. This case is impossible, because theproblem stipulates that r, s are nonzero.

Finally, our answer is 8 + 20 + 52 = 80 .

The above problem was hardly difficult at all - we just needed to keepour head, avoid the numerous pitfalls (such as not recognizing s must bepositive for example), and keep our work organized. Do this, and you’vesolved an AIME problem 10 in just minutes.

Problem 7.0.42. Let S be the set of all perfect squares whose rightmostthree digits in base 10 are 256. Let T be the set of all numbers of the formx−2561000

, where x is in S. In other words, T is the set of numbers that resultwhen the last three digits of each number in S are truncated. Find theremainder when the tenth smallest element of T is divided by 1000. (2012AIME I Problem 10)


Solution. This is another problem that really just requires keeping our headand staying organized. Let’s call k an element of S. Then we have

k2 ≡ 256 (mod 1000)

=⇒ (k − 16)(k + 16) ≡ 0 (mod 1000)

It may be tempting to claim that k − 16 ≡ 0 (mod 1000) or k + 16 ≡ 0(mod 1000), but this is not the case. It would indeed be true if k − 16 andk + 16 were relatively prime, but not otherwise.

Notice that 1000 = 23 · 53. Furthermore, if k− 16 is a multiple of 5, thenk + 16 is not - and vice versa. This implies that either k − 16 is a multipleof 125 or k + 16 is. We can apply similar (but not the same!) logic to thepowers of 2. Since k−16 ≡ k+16 (mod 4), we must have 4 dividing k−16.

As such either 500 divides k − 16 or 500 divides k + 16, meaning thatk ≡ ±16 (mod 500). Therefore, the smallest element of S is 162, the secondsmallest is (500− 16)2, followed by (500 + 16)2, then (500 · 2− 16)2, and soon. Thus, the 10th smallest number is x = (500 · 5− 16)2 = (2500− 16)2 =25002−2·16·2500+162, meaning that x−162

1000= 25002−2·16·2500

1000= 252·10−16·5 =

6250− 80 = 6170, making the answer 170 .

This was another fairly straightforward problem, and in fact one thatwas very easy to “see” (or guess) the answer to. Notice how we indirectlyused the following fact:

Theorem 7.0.28. If p is a prime number, and p divides the product ab,then either p divides a or p divides b.

This is, of course, immediately obvious, but still incredibly important.Most likely you have used this fact without even realizing it at some pointin your life.

Problem 7.0.43. Find the sum of all positive integers n with no more than3 digits for which the number obtained as the last three digits of n2 equalsn. (Lehigh 2014 Problem 38)

Solution. This one also isn’t too bad. The condition tells us that n2 ≡ n(mod 1000), implying that n ≡ 1 (mod 1000

gcd(1000,n)) (don’t forget how modu-

lar division works!). Notice that 1000 = 23 ·53. Let’s say that gcd(1000, n) =k. This implies, of course, that k | n. But it also implies that n ≡ 1


(mod 1000k

). Furthermore, since k | 1000, k is of the form 2a · 5b for somea, b ≤ 3.

If a is 1 or 2, then this would simultaneously imply that 2 | n (as 2 | 2a |k | n) and n ≡ 1 (mod 2) (as 2 | 1000

k), which is clearly absurd. Thus a = 3

or a = 0. By similar reasoning, either b = 0 or b = 3.Let’s look at these four cases in turn. If a = b = 0 then we have n ≡ 1

(mod 1000). As n has at most 3 digits, this gives the only solution n = 1. Ifa = 0 and b = 3, then we have n ≡ 1 (mod 8) and 125 | n, giving the singlethree-digit solution 625. If a = 3 and b = 0, then we have n ≡ 1 (mod 125)and 8 | n, giving the single solution 376. If a = b = 3, then 1000 | n, whichis not possible.

Therefore, our answer is 1 + 376 + 625 = 1002 .

Notice how we just went through three supposedly very difficult problems(two #10s on the AIME and a #38 (of 40) on the Lehigh test), but inreality they were simple applications of writing out the modular arithmeticequations and being very careful about solving them. Especially underthe pressure of a contest, it is very easy to make a silly mistake and ruinan entire question. For example, many people answered 160 to the firstquestion, overlooking the fact that (c, r, s) and (c, r,−s) lead to the samesolution. Similarly, several people forgot to count 162 in the second solution,and on the third question I erroneously concluded that 8 | 126 (severalothers forgot the solution n = 1). In actuality, this holds in general formost high-numbered problems. Although this can be very frustrating, youshould be very happy about this! With few exceptions, there’s no heavymachinery involved, and all of the concepts are simple enough to explain toa relatively young child. All that is necessary is to stay calm and keep yourwork organized, assuming that you know the basics (which, if you read thefirst issue, you should!).

Problem 7.0.44. Let R be the set of all possible remainders when a numberof the form 2n, n a nonnegative integer, is divided by 1000. Let S be thesum of all elements in R. Find the remainder when S is divided by 1000.

Solution. The previous article proves useful! Clearly, at some point, thepowers of 2 are going to repeat modulo 1000 (there’s only so many re-mainders modulo 1000!), so we are looking for a k such that 2n+k ≡ 2n

(mod 1000). Sound familiar?We have to be very careful here. Notice that 20 = 1, 21 = 2, and 22 = 4.

But for 23, 24, and beyond, these remainders are never going to be seen


again (as these are all multiples of 8). So in the above equation, 2n+k ≡ 2n

(mod 1000), we can assume that n ≥ 3. This allows us to divide through by2n, giving us 2k ≡ 1 (mod 125).

We already know how to do that! We have φ(125) = 100, meaning that2100 ≡ 1 (mod 125). However, this does not necessarily mean we’re done. Itis possible that there is a smaller k such that 2k ≡ 1 (mod 125). Withoutgoing into too much detail here, we can pretty quickly verify that this isnot the case with some quick computation (it suffices to check that 220 6≡ 1(mod 125) and 250 6≡ 1 (mod 125), but this is a discussion for another day),so indeed 100 is the smallest such k.

Thus, the possible remainders are 20, 21, 22, . . . , 2102. Adding these givesus 2103 − 1, but we already know that 2103 ≡ 23 (mod 1000). Thus, ouranswer is 8− 1 = 007 .

So why were we able to skate over the order details? Let’s look at thesmallest k such that 2k ≡ 1 (mod 125). Then the possible remainders wouldbe 20, 21, . . . , 2k+2, since 2k+3 ≡ 23 (mod 1000). Their sum would be 2k+3−1,but we already established that 2k+3 ≡ 23 (mod 1000) - making our answer007 regardless! Such tricks are often useful, even when they don’t give muchinsight into what’s really going on in the question. On a contest such as theAIME, your goal is to maximize your score - not the number of problemsyou solve completely.

Problem 7.0.45. For a positive integer p, define the positive integer n tobe p-safe if n differs in absolute value by more than 2 from all multiples of p.For example, the set of 10-safe numbers is 3, 4, 5, 6, 7, 13, 14, 15, 16, 17, 23, ....Find the number of positive integers less than or equal to 10, 000 which aresimultaneously 7-safe, 11-safe, and 13-safe.(2012 AIME II Problem 12)

Solution. This one, once you get past the notation, is very simple. A numberis 7-safe if it is equivalent to 3 or 4 in modulo 7, 11-safe if it is equivalentto 3,4,5,6,7, or 8 modulo 11, and 13-safe if it is equivalent to 3, 4, 5, 6, 7,8, 9, or 10 modulo 13. Since each of these cases map to a unique residuemodulo 1001 by the Chinese Remainder Theorem (or pure intution, whichwill do fine here), there are 96 such residues. As such, there are exactly 960numbers that are 7-safe, 11-safe, and 13-safe that are less than 10010.

However, we need to subtract off the numbers that are greater than10000. Fortunately, we can easily check these numbers - only 10006 and10007 are simultaneously 7-safe, 11-safe, and 13-safe. Therefore, our finalanswer is just 960− 2 = 958 .


Don’t be scared by the notation above! Although it may appear likewe killed this problem by citing some obscure theorem, in reality we didnot. Once again, we simply wrote out the given information, and used basiccommon sense to finish up the problem. The Chinese Remainder Theorembit is just to formalize our intution, but it should be relatively obvious thatfor any a1, a2, a3, there exists a unique residue x modulo 7 · 11 · 13 suchthat n ≡ x (mod 1001) simultaneously satisfies n ≡ a1 (mod 7), n ≡ a2(mod 11), and n ≡ a3 (mod 13).

The final problem is really an embodyment of what we’ve been talkingabout this entire time. It uses absolutely no fancy tricks whatsoever, isreally just a matter of synthesizing the given information into easily workableequations (which, frankly, is not too difficult), and is easy to solve as long aswe avoid some basic silly errors. This should sound familiar - it’s essentiallyalmost all of the problems we’ve done so far. Although we have cited some“fancy” theorems such as FLT, Euler’s Theorem, and the Chinese RemainderTheorem, this was solely to construct a rigorous proof and is hardly necessaryto do on an actual competition. Notice, however, that this final problemis problem 15 - the last problem on the test (and thus, theoretically, thehardest).

Problem 7.0.46. There are n mathematicians seated around a circulartable with n seats numbered 1, 2, 3, · · · , n in clockwise order. After a breakthey again sit around the table. The mathematicians note that there is apositive integer a such that

(1) for each k, the mathematician who was seated in seat k before thebreak is seated in seat ka after the break (where seat i+ n is seat i); (2) forevery pair of mathematicians, the number of mathematicians sitting betweenthem after the break, counting in both the clockwise and the counterclock-wise directions, is different from either of the number of mathematicianssitting between them before the break.

Find the number of possible values of n with 1 < n < 1000. (2012 AIMEI Problem 15)

Solution. The first step is to decipher the language. The first condition istelling us that, for every x, y ∈ {1, 2, 3, . . . , n}, that ax 6≡ ay (mod n). In

other words, x 6≡ y (mod)

( ngcd(n,a)

). This need not be scary - because x, y

range over all pairs 1 ≤ x, y ≤ n (where x, y are distinct), this implies thatgcd(n, a) must be 1 - or a is relatively prime to n.

The second condition says that x−y 6≡ (mod a)x−ay (mod n) and x−y 6≡ (mod a)y−ax (mod n), since x−y−1 is the number of mathematicians


sitting between them before the break and ±(ax − ay) − 1 is the numberafterwards. As such, we have (x−y)(a−1) 6≡ 0 (mod n) and (x−y)(a+1) 6≡0 (mod n). As, again, x, y range over all possible pairs, this implies thata− 1 and a+ 1 are also relatively prime to n.

Thus, our answer is simply the number of n < 1000 such that there existsan a with a− 1, a, a+ 1 all relatively prime to n. If 2 | n or 3 | n, there doesnot exist such an a by the Pigeonhole Principle (or just sheer common sense- among any k consecutive integers there is a multiple of k), so our answeris simply the number of n with 1 < n < 1000 that are not multiples of 2 or3.

This is simply (Number of integers in the set {2, 3, . . . , 999})-(Number ofmultiples of 2 in the set {2, 3, . . . , 999})-(Number of multiples of 3 in the set{2, 3, . . . , 999})+(Number of multiples of 6 in the set {2, 3, . . . , 999}) - noticethat we add on the multiples of 6 to avoid overcounting. This evaluates to998− 499− 333 + 166 = 332 .

Amazing! Once again, all we really did was keep our head, set up ourequations properly, and make some pretty simple observations. Nothingfancy was required, no clever manipulations, just some common sense (andthe Principle of Inclusion-Exclusion at the end, but this is not particularlydifficult). Despite this, we easily took apart what is supposed to be the mostdifficult question on the test.

Summary:In this article, we saw various problems in Number Theory, spanning lit-

erally from the beginning of the AIME to the end of the test. Although all ofthese problems look very different on the surface, each of them is approachedin more or less the same manner. We simply set up our modular equations,and be very careful in working with them. Of course, this is often easiersaid than done, but there is rarely any particularly clever manipulation orheavy machinery involved. Instead, we can (and did! Reread N2 in the firstissue if none of this made sense whatosever) summarize the necessary back-ground in just a few pages, and be able to apply them to problems almostimmediately. As we saw, even #15s fall to this method without any real pain.

Exercises:

1. Find the smallest positive integer x such that

• x is 1 more than a multiple of 3




• x is 9 more than a multiple of 11 and

• x is 2 more than a multiple of 13.

(PUMaC 2013)

2. Determine the smallest positive integer that is both three times asquare and two times a cube.

3. Find all prime p such that 5p + 12p is a perfect square.

4. Kelvin the frog’s relatives are coming over for a party. Altogether,they make up 201420142014 frogs. With so many frogs, inevitibly thereare many in-law relationships and so frogs gradually drift into groups.Astoundingly, every single one of these groups has exactly 110 frogs,except for Kelvin the frog’s smaller group which has only n frogs (in-cluding Kelvin). Compute n.

5. What is the smallest positive integer n such that 2013n ends in 001(i.e. the rightmost three digits of 2013n are 001?) (PUMaC 2013)

6. The set of positive integers is arranged in an infinite grid. The numberone is placed in the top left in the first diagonal, then in the kthdiagonal running from lower left to upper right, the next k numbersare placed in order starting at the lower left. The first few rows are

1 3 6 10 15 . . .2 5 9 14 . . .4 8 13 . . .7 12 . . .11 . . .. . .

Let f(n) = xy, where x is the row that n is in and y is the columnthat n is in. For example, f(5) = 4 and f(9) = 6. Compute f(2014).

7. Let ak denote the sum 1! + 2! + 3! + . . . + k!, and let bk denote thesum a1 + a2 + . . . + ak. For how many n ≤ 2013 does bn end in a3?(djmathman’s mock AMC 10 question 18)


8. A number n is considered to be k-seperable if there exists an integera such that n = a(a+ k). Find the sum of all numbers which are both2-seperable and 9-seperable (djmathman’s mock AMC 10 question 23)

9. Let rad(n) be the product of the distinct prime factors of n. For exam-ple, rad(30) = rad(60) = rad(120) = . . . = 30. Note that rad(1) = 1.Furthermore φ(n) is the number of positive integers less than or equalto n that are relatively prime to n. Then the infinite sum

∞∑a=1

∞∑b=1

∞∑c=1

rad(2a3b5c)φ(2a3b5c)

(2a3b5c)3

can be written as mn

, where m and n are relatively prime positiveintegers (if the answer is an integer, take n = 1). Compute m+ n.

10. Let N be the set of natural numbers {1, 2, 3, . . .}. Define a functionf : N → Z by f(1) = 0, f(p) = 1 for any prime p, and f(mn) =mf(n) + nf(m) for any integers m,n. Find all n such that f(n) = n.

8

NEW! Contests and monthlyleaderboards

Starting with this issue, QEDMonthly is launching a new intiative to stim-ulate activity within the magazine. Every month, we release a great deal ofexercises, most of which go unsolved and many of which go untried. Thisis especially distressing in the lower sections of levels, where not a singleproblem had a solution posted. To combat this, we are announcing a newpoint-based system, complete with fancy prizes!

Each month, with the release of the new issue, the QEDMonthly authorswill go back over the last month’s problems and select what we believe to bethe best solution to each problem. The author of that solution will receiveone point, meaning that in this particular issue 80 points are up for grabs.Solutions will be chosen on a few criteria:

• Correctness. Obviously, to get points, your solution should be correct.

• Applicability to the article. If the article is teaching some topic andyou solve the problem in a completely different method, this will hurtyour chances of receiving points.

• Well-written/Clear. The solution should be easy to follow and notmake any apparently sudden leaps, even if they are true.

• Instructive. Writing “this is a trivial application of Brokard’s theo-rem”, even if completely correct, is completely useless. The level ofthe solution should match the level of the problem.

There are also several other ways to earn points. Submitting an articlethat ends up being published (with editing of course) can earn upwards of

87


20 points, depending on how much editing was necessary. Submitting aproblem that ends up as a contest problem is also worth points. The exactpoint values are not yet set, but will be shortly announced. Suffice it to saythat the more you do, the more points you’ll get!

Every 3 months, the leaderboard will reset and prizes will be awarded.The exact prize system is not yet determined, but the prizes will likely beT-shirts and the other usual suspects.

Additionally, beginning with the release of issue 3, QEDMonthly will berunning its very own contest! The contest format is tentative, but unlikelyto change significantly. Every 6 months, a new contest cycle will begin.Each contest cycle will consist of four major rounds. The first round willconsist of several problems around the level of the introductory articles,and the top participants will qualify to round 2 (the exact number dependson participation). This round will only require answers - proofs are notrequired. The second round will feature several problems around the levelof the intermediate articles, combining a mix of short-answer and proofproblems. Again, the top participants will advance on to round 3. This willbe an especially challenging round, featuring problems around the level ofthe advanced articles, including some olympiad-level problems. Full proofswill be expected on each problem.

The top few participants on round 3 will be invited to take part in the“live” competition, to be held online at some mutually convenient time.Spectating will be allowed and highly encouraged, as this round is sure tobe exciting. Modeled after MATHCOUNTS countdown, the participantswill face off head-to-head in a fast-paced quizbowl type competition, withthe winners receiving prizes (again, probably T-shirts and such). Theseproblems will be around the mid-AMC level, or somewhere around the earlyparts of the intermediate articles. Again, full details will be released at theactual time.

Hopefully, these new intiatives will facilitate both of the goals of QED-Monthly - to introduce students to the world of math competitions in sucha method that will leave them thirsty to continue, and to help studentsalready in math competitions along each step of their quest towards theolympiads. The competitive part is secondary, but because competition isvery good at stimulating interest we are experimenting with these designs.As always, you are welcome to leave comments, suggestions, and feedbackon the appropriate AoPS thread.

9

Credits and Mock AIME Information

I owe a great deal of gratitude to various sources for making this possible,whether directly or indirectly. Many years ago I was extremely fortunateto come across the Art of Problem Solving, whose various classes, forums,and the like caused math to draw me in, something that would undoubtedlynever have happened otherwise. We also would like to thank math com-petitions worldwide for providing an incredibly effective method of learningmathematics - you have undoubtedly been the catalyst that has caused thou-sands of students to decide they wanted to do math. We also thank themfor the numerous high-quality problems that they write, many of which havebeen used as instructional examples or exercises in this issue.

Again, we would like to thank all the users associated with the Mathtimeproject (which are far too many to name!), which provided the inspirationfor this magazine. Without their vision, I likely would never have evenhad a stray thought about creating QEDMonthly. As usual, we would liketo thank everyone on the AoPS forums who directly or indirectly providedmaterial for this magazine, and especially those users that provided helpfulsuggestions and feedback after the first issue.

And, of course, we thank all the readers of this magazine. If nobody readthis magazine, there wouldn’t be much point in producing it! We hope thatall (or least some) of you find these productions instructive, helpful, butmost of all clear, natural, and enjoyable. We also hope that you will aid usin spreading the word about QEDMonthly. Tell your math team about us,or show them some snippets from our articles. Discuss the problems withyour team. Tell your friends, even if you don’t think they’d be interestedat all. The goal is to have as much people benefit from QEDMonthly aspossible, but someone can’t benefit from it very much if they have no idea

89


it even exists!All material in this issue, besides the obvious exceptions of problems and

exercises drawn from various sources, was written by Alexander Katz. JustinStevens provided the LATEXtemplate that has been used for both issues so far,for which (along with his articles!) we thank him profusely. Kelvin Wang,who may be better known as Kelvin the Frog among our readers, has joinedthe QEDMonthly team for the future and provided helpful suggestions andfeedback in molding the competitive side of QEDMonthly.

On the mock AIME front - Many of you have likely already seen thisannouncement, but I have written a full mock AIME which will be availableon the AoPS forums (AoPS =⇒ Community =⇒ American MathematicsCompetitions =⇒ “Kelvin the Frog returns!”) this weekend (either late3/8 or early 3/9). This is intended to be a complete replica of the AIME,featuring 15 short-answer problems to be solved in 3 hours. More detailsare available in the thread.

If you are taking this mock AIME as a result of reading this, pleaseindicate “QEDMonthly2014” somewhere in your submission - this will makeyour score eligible for some bonus points in our new scoring system! Theexact scaling factor is not yet determined, but free points are always good:) To receive credit, your submission must be received before the AIME I,meaning that it should be received by March 12th.

A huge thanks to all of you for reading this, and we hope you enjoy!

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