Topic 1: Algebra Dr J Frost ([email protected])

Topic 1: Algebra

Dr J Frost ([email protected])

Slide guidanceKey to question types:

IMC Senior Maths Challenge

BMO British Maths Olympiad

The level, 1 being the easiest, 5 the hardest, will be indicated.

Those with high scores in the IMC qualify for the BMO Round 1. The top hundred students from this go through to BMO Round 2.Questions in these slides will have their round indicated.

Frost A Frosty SpecialQuestions from the deep dark recesses of my head.

Classic ClassicWell known problems in maths.

? Any box with a ? can be clicked to reveal the answer (this works particularly well with interactive whiteboards!).Make sure you’re viewing the slides in slideshow mode.

A: London

For multiple choice questions (e.g. IMC), click your choice to check your answer (try below!)

Question: The capital of Spain is:

B: Paris C: Madrid

Tip #1: Being adept with laws of indices

Level 2

Level 1

Level 4

Level 5

Level 3

IMC

You need to be well practiced with your laws of indices – many IMC questions in particular rely on this skill.

Question: What is half of 220?

A: 110

B: 120 C: 20

D: 210 E: 219

Half of 220 is: 220 / 2 = 220 / 21 = 219


Level 2

Level 1

Level 4

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IMC


Question: Given that 4x + 4x + 4x + 4x = 416, what is the value of x?

You can discuss this question in groups/pairs. Raise your hand when you’ve got an answer.

A: 2

B: 4 C: 8

D: 12 E: 15

3√(212) = 24

Tip #1: Being adept with laws of indicesLet’s practice our laws of indices. Simplify the following.

2 · 2x = 2x+1

((x3)3)3 = x27

This dot means multiplication.

2x / 2 = 2x-1

√(210) = 25

2x · 4y = 2x+2y?

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9x · 3x = 33x = 27x

Hint: make the base of the second power expression 2.

In general, ac x bc = (ab)c

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2x + 2x + 2x + 2x = 2x+2

4x + 4x + 4x + 4x = 4x+1

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Therefore on the previous slide 4x+1 = 416, and so x+1 = 16, and so x = 15.

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Level 2

Level 1

Level 5

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IMC


Question: Below are three statements. Exactly which ones are true? (i) 310 is even (ii) 310 is odd (iii) 310 is square

A: (i) only

B: (ii) only C: (iii) only

D: (i) + (iii) E: (ii) + (iii)

310 is square because its square root is 35.An odd number multiplied by an odd number is odd, so 3x is always odd for (positive) x.


Level 2

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IMC


Question: Given that 5p = 9, 9q = 12, 12r = 16, 16s = 20 and 20t = 25, what is the value of pqrst?

A: 1

B: 2 C: 3

D: 4 E: 5

What might be going through your head: “Hmm, it seems as if the expressions are ‘chained’ together. And since I’m looking for pqrst, it indeed looks like I’ll have to combine them together.”

If 5p = 9 and 9q = 12, then (5p)q = 12. Continuing with this, we get ((((5p)q)r)s)t = 25, so 5pqrst = 25, and so pqrst = 2.

Tip #1: Being adept with laws of indicesSummary of tips:

When a number is on its own, its implicit power is 1. This can help us simplify expressions.

2 · 2x = 21 · 2x = 2x+1

Making numbers the same base again helps us simplify.

4x · 2y = (22)x · 2y = 22x · 2y = 22x+y

Taking the square root halves the power.√(212) = √(26)2 = 26

Tip #2: ?

Question: Simplify 1 – x2

(1+x)(1-x)

Same algebraic questions rely on you identifying and exploiting the

difference of two squares. Make sure you can spot it!

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Tip #2: Using the difference of two squares

Level 2

Level 1

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IMC

Question: Which of the following is equal to (1 + x + y)2 – (1 – x – y)2 for all values of x and y?

A: 4x

B: 2(x2+y2) C: 0

D: 4xy E: 4(x+y)

Note: There would be nothing stopping you expanding the original expression then simplifying, but it would waste time.

By using the difference of two squares to factorise, we get (2)(2x + 2y) = 4(x+y).

Tip #2: Using the difference of two squares

Level 2

Level 1

Level 4

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Level 5

IMC

Question: What is the largest power of 2 that divides 1272 − 1?

A: 21

B: 27 C: 28

D: 263 E: 2127

Using the difference of two squares, we get: 126 x 128. At this point, we look at how many factors of 2 are in each.We can divide 126 by 2 just once before we get to an odd number, but 128 is 27. So in total, we have 8 factors of 2.

Tip #3: More difficult simultaneous equationsIn the past you’ve solved simultaneous equations like 2x + 3y = 4 and x + 6y = 7. You may have solved these by scaling and then adding or subtracting the equations.But for more complicated equations, we can use substitution.

x2 + y2 = 12 – 4xx – y = 2

Step 1: In the simpler equation, make one of the variables the subject of the formula. e.g. y = x - 2

Step 2: This allows us to get rid of y in the first equation by substituting y with x – 2.

x2 + (x-2)2 = 12 – 4x

x2 + x2 – 4x + 4 = 12 – 4x

Step 3: Expand, simplify and solve as you usually would.

2x2 = 8 x2 = 4

x = 2

Tip #3: More difficult simultaneous equations

Cayley

Hamilton

Macclaurin

IMO

Question: Find all real values of and y that satisfy the equations:

x4 – y4 = 5x + y = 1

Use what you know!

That first equation looks suspiciously like the difference of two squares...

I know how to simplify and substitute.

In the past you’ve solved simultaneous equations like 2x + 3y = 4 and x + 6y = 7. For

Let’s break this up into 2 steps:

What’s the simplest form we can put the first equation in?

Answer: (x-y)(x2 + y2) = 5?

(x4 – y4) = (x2 - y2)(x2 + y2) = (x-y)(x+y)(x2 + y2) = (x-y)(x2 + y2) because x + y = 1 from the second equation.

Tip #3: More difficult simultaneous equations

Cayley

Hamilton

Macclaurin

IMO

Question: Find all real values of and y that satisfy the equations:

x4 – y4 = 5x + y = 1

We need to get rid of y for the moment. y = 1 – x from the second equation. So substituting that into our simplified from of the first equation, we get:(2x – 1)(x2 + (1-x)2) = (2x – 1)(2x2 – 2x + 1) = 5.Expanding and simplifying gives us 2x3 – 3x2 + 2x – 3 = 0.Usually we don’t know how to factorise cubics, but it’s quite easy in this particular case:x2(2x – 3) + 1(2x – 3) = 0. So (x2 + 1)(2x – 3) = 0.Either x2 + 1 = 0 (which doesn’t have a real solution) or 2x – 3 = 0, giving x = 3/2. And since y = 1 – x, y = -1/2.

Use what you know!

That first formula looks suspiciously like the difference of two squares...

I know how to simplify and substitute.

In the past you’ve solved simultaneous equations like 2x + 3y = 4 and x + 6y = 7. For

Now let’s do the substitution...

Tip #4: Introducing variables

Macclaurin

Hamilton

Cayley

IMO

Question: Before the last of a series of tests, Sam calculated that a mark of 17 would enable her to average 80 over the series, but that a mark of 92 would raise her average mark over the series to 85. How many tests were in the series?

Sometimes we have to introduce variables of our choosing to help model (and subsequently solve) a problem.

Answer: 15

Use a variable x to represent the total from the previous papers, and n the number of papers. The using the information provided:

This gives us x + 17 = 80n and x + 92 = 85n. These are simultaneous equations! Subtracting the first from the second gives 75 = 5n, so n = 25.

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Use what you know!

I know the formula for the mean average.

I know how to form an expression using given information.

Since I’ll get two equations, I know how to solve simultaneous equations.

Tip #5: Dealing with ratios

Consider the lengths x and y of two planks of wood. The ratio of their lengths is 2:3. How could we form an equation?

In general, if the ratio in size of x and y is a:b, then bx = ay. Don’t get this the wrong way round!

Question: The area of shapes A and B are x2 and 3x – 4. The ratio of their area is 2:1. Find the possible values of x. [Source: Frosty Special]

x

y

If we had 3 of the first plank, that would be the same length as 2 of the second plank. So 3x = 2y.

Answer: 2 or 4.?1(x2) = 2(3x-4). Simplifying and rearranging, this gives us x2 – 6x + 8 = 0.Factorising, (x-4)(x-2) = 0, so x = 4 or x = 2.

Tip #5: Dealing with ratios

Macclaurin

Hamilton

Cayley

IMO

Question: A rectangular piece of paper is cut into two shapes by a straight line passing through one corner, as shown. Given that area X : area Y = 2:7, what is the ratio of a : b?

Answer: 4:5

Let’s represent the width of the rectangle by a variable w.Area of triangle: ½ waArea of triangle: ½ ([a+b] + b)w = ½ (a + 2b)wSo using the ratios, 7(½ wa) = 2(½ (a + 2b)w)Cancelling an simplifying: 7a = 2a + 4b, so 5a = 4b.Using our ‘ratio tip’ backwards, the ratio of a to b is 4:5.By checking the diagram, this answer looks sensible.

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Use what you know!

I know the area of a triangle and trapezium.

I can deal with ratios to form an equation.

I can introduce variables to represent unknown quantities.

I can solve equations that I have formed.

X

Y

a

b

Tip #6: Cancelling/simplifying intermediate terms

Level 2

Level 1

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IMC

Question: Given that the number 2006 is the correct answer to the calculation 1 − 2 + 3 − 4 + 5 − 6 + … + (n − 2) − (n − 1) + n, what is the sum of the digits of n?

In some algebraic problems, either middle terms in a sum/multiplication cancel often leaving the ‘ends’, or we can easily simplify them in some way.

What happens if we group the numbers in pairs after the first term?

Advice:

-2+3 = 1. So is -4+5. Therefore the entire sum is 1 + 1 + 1 + 1 + ... + 1. How many ones are there?

If we exclude the first term there’s n-1 terms. So there’s (n-1)/2 pairs. If we include the initial 1, that’s 1 + (n-1)/2 ones we have (which could simplify to (n+1)/2 ).

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Level 2

Level 1

Level 4

Level 3

Level 5

IMC

Question: Given that the number 2006 is the correct answer to the calculation 1 − 2 + 3 − 4 + 5 − 6 + … + (n − 2) − (n − 1) + n, what is the sum of the digits of n?

A: 3

B: 4 C: 5

D: 6 E: 7


We found that the above expression simplifies to (n+1)/2. What therefore is the answer?

(n+1)/2 = 2006, so solving we get n = 4011. The digits add up to 6.


Frosty Special: Simplify the following.


Let’s go through a potential strategy together.

What will each of the brackets be as an improper fraction?

13

24

35

n-4n-2

n-3n-1

n-2n

...? ? ?

? ?

?

?

?

?

?

Tip #6: Cancelling/simplifying intermediate termsIn some algebraic problems, either middle terms in a sum/multiplication cancel often leaving the ‘ends’, or we can easily simplify them in some way.

Next job: What can we cancel in this multiplication?

13

24

35

n-4n-2

n-3n-1

n-2n

...46

These diagonals are the same.

As are these.Then we consider what we’d have left if we cancelled all these terms.

__2__n(n-1)

Final answer: ?

Topic 1: Algebra Dr J Frost ([email protected])

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Transcript of Topic 1: Algebra Dr J Frost ([email protected])