# Maths Lab: Wordings - · PDF fileMaths Lab: Elements of solution Singapore Mathematics...

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Maths Lab: Wordings

Lfs Maths Lab, SMO Senior 1

Maths Lab: Elements of solutionSingapore Mathematics OlympiadSenior Section 2010Duration: 2 h30.

The number of indicates the relativ difficulty of the question.

Multiple Choice Questions

1 Answer (A).

Find the value of H123L + H246L + H369L + + H3356701005L

H136L + H2612L + H3918L + + H33510052010L .We have

H123L + H246L + H369L + + H3356701005LH136L + H2612L + H3918L + + H33510052010L =

H123LA1 + 23 + 33 + + 3353EH136LA1 + 23 + 33 + + 3353E then

H123L + H246L + H369L + + H3356701005LH136L + H2612L + H3918L + + H33510052010L =

1

3.

2 Answer (E).

Lets a, b, c and d be real numbers such that b + c + d

a=

a + c + d

b=

a + b + d

c=

a + b + c

d= r .

As r + 1 = 1 +b + c + d

a=

a + b + c + d

a so aHr + 1L = a + b + c + d .

Identically, bHr + 1L = cHr + 1L = d Hr + 1L = a + b + c + d so Hr + 1L Ha + b + c + d L = 4 Ha + b + c + d L.Thus we obtain Hr - 3L Ha + b + c + d L = 0.Then r = 3 or a + b + c + d = 0.

Yet a + b + c + d = 0 gives -a = b + c + d so b + c + d

a=-a

a= -1 and identically -b = a + c + d , -c = a + b + d and

-d = a + b + c so a + c + d

b=

a + b + d

c=

a + b + c

d= -1.

Finally, r = 3 or r = -1.

3 Answer (E).

If 0 < x 0.As 4907 = 7701 with 701 prime number.

(In fact as 252 < 701 < 302, we know that if 701 wasnt a prime number it will be divisible by 2,3,5,7,11,13,17,19,23 or29 and thats not the case.)Therefore we obtain:

k - m = 1 and k + m = 4907, then k = 2454 and n = 2453 k - m = 4907 and k + m = 1, then k = 2454 and n = -2453 k - m = 7 and k + m = 701, then k = 354 and n = 347 k - m = 701 and k + m = 7, then k = 354 and n = -347 .

As y = n2, we deduce y = 24532 or y = 3472.Moreover

or x - 90 = 2454 so x = 2544 or 90 - x = 2454 so x = -2364: we have 2 ordered pairs I-2364, 24532M andI2544, 24532M,

or x - 90 = 354 so x = 444 or 90 - x = 354 so x = -264: we have 2 ordered pairs I-264, 3472M andI354, 3472M .

7 Answer (D).

Lets describe the cases of non-empty subsets verifying the condition Good: the singletons: only composed by an even number so 5 subsets;

the pairs: two even numbers or 1 even and 1 odd, so 52

+51

51

= 10 + 25 = 35 subsets;

the 3lists: two even and one odd or three even, so 52

51

+53

= 105 + 10 = 60 subsets;

the 4lists: 2 even and 2 odd, 3 even and 1 odd or 4 even, so 52

52

+53

51

+54

= 100 + 50 + 5 = 155

subsets;

the 5lists: 3 even and 2 odd, 4 even and 1 odd or 5 even, so 53

52

+54

51

+55

= 100 + 25 + 1 = 126

subsets; the 6lists: 3 even and 3 odd, 4 even and 2 odd or 5 even and 1 odd, so

53

53

+54

52

+55

51

= 100 + 50 + 5 = 155 subsets;

the 7lists: 4 even and 3 odd or 5 even and 2 odd, so 54

53

+55

52

= 50 + 10 = 60 subsets;

the 8lists: 4 even and 4 odd or 5 even and 3 odd, so 54

54

+55

53

= 25 + 10 = 35 subsets;

the 9lists: 5 even and 4 odd so 55

54

= 5 subsets;

the 10lists: 5 even and 5 odd so 1 subset.the overall number of subsets is then 5 + 35 + 60 + 155 + 126 + 155 + 60 + 35 + 5 + 1 = 637 subsets.

8 Answer (B).

As f Hr L = k and f HsL = k, then r and s are symetric regarding the x-coordinate of the summit of the quadratic function,it is to say -

b

2 a.

We then deduce s + r = 2 -b

2 a= -

b

a .

Thus f Hr + sL = a - ba

2

+ b -b

a+ c =

b2

a-

b2

a+ c = c .

4 Lfs Maths Lab, SMO Senior

9 Answer (D).

Find the number of positive integers k < 100 such that 2 H36 nL + kH23 n+1L - 1 is divisible by 7 for any positive integer n.Note that 2 H36 nL + kH23 n+1L - 1 = 2 H36Ln + 2 k H23Ln - 1 = 2 HH36Ln + k H23LnL - 1.We have 23 = 8 = 7 + 1 so the remainder of 23 in the division by 7 is 1. Thus the remainder of 23 n is 1 too.

Identically, 36 = H32L3 so the remainder of 36 in the division by 7 is 1, thus the remainder of 36 n is 1.Finally the remainder of HH36Ln + k H23LnL is 1 + k1 = k + 1.We have to determine the number of positive integers k < 100 such that 2 Hk + 1L - 1 = 2 k + 1 is divisible by 7.So it must exist integer p such that 2 k + 1 = 7 p , 2 k = 7 p - 1 = 7 H p - 1L + 6 and we deduce that k = 7 p ' + 3.As 0 7 p ' + 3 < 100, 0 p ' < 13 .We obtain 14 possibilities.

10 Answer (C).

Lets consider point E, intersection of the lines AD et CM.The angles A M E and B M E are opposite angles they are congruent.As lines AD and CB are parallel and line CM is a secant then the angles A E M and B C M are alternate angles andcongruent.Moreover A M = M E as M is the midpoint of AB .Theregore the triangles AEM and CBM are congruent.We deduce A E = C B and M E = C M so A E + A D + D C = 17 and C E = 13 and clearly the area of the trapeziumis equal to the area of triangle CDE.The triangle EDC is rectangle at point D, according to the Pythagoreas theorem, C E2 = C D2 + D E2.

So 132 = C D2 + H17 - D CL2 so D C is solution of the equation 2 x2 - 34 x + 120 = 0.The roots of this quadratic equation are 5 and 12.

(The discriminant is D = 196 > 0 so the roots are 34 - 196

22= 5 and

34 + 196

22= 12).

Finally D E = 12 or D E = 5 so the area of the triangle is in both cases 512

2= 30.

Remark:Lets note S the area of trapezium ABCD.

Then S =1

2D ED C as previously explained.

Now HD E + D CL2 = D E2 + 2 D ED C + D C2 = 4 S + C E2. In fact, Pythagoreas Theorem in triangle DEC right-angled at D, shows that C E2 = D E2 + D C2.

Finally as D E + D C = A D + C D + C B = 17, we deduce 4 S + 132 = 172 and so S = 30 .

Short Questions

11 Answer: x = 132 .

Denote L et w the dimensions of the original rectangle.Then the area is given by A = Lw .We then know that HL + 6L Hw - 2L = Lw so we obtain -2 L + 6 w - 12 = 0, L - 3 w + 6 = 0.We also know that HL - 12L Hw + 6L = Lw so we also obtain 6 L - 12 w - 72 = 0 , L - 2 w - 12 = 0.So the dimensions of the original rectangle are the couple solution of the system

L - 3 w + 6 = 0L - 2 w - 12 = 0

.

Thus L = 48 and w = 18 so x = 2 HL + wL = 132.

Lfs Maths Lab, SMO Senior 5

Denote L et w the dimensions of the original rectangle.Then the area is given by A = Lw .We then know that HL + 6L Hw - 2L = Lw so we obtain -2 L + 6 w - 12 = 0, L - 3 w + 6 = 0.We also know that HL - 12L Hw + 6L = Lw so we also obtain 6 L - 12 w - 72 = 0 , L - 2 w - 12 = 0.So the dimensions of the original rectangle are the couple solution of the system

L - 3 w + 6 = 0L - 2 w - 12 = 0

.

Thus L = 48 and w = 18 so x = 2 HL + wL = 132.

12 Answer: 2575.

For all integer r, u r = 1 + 2 + + r =r Hr + 1L

2.

Thus for all integer j 1, i=1

j 1

u i=i=1

j 2

iHi + 1L .

Lets remind that 1

i-

1

i + 1=

i + 1 - i

iHi + 1L =1

iHi + 1L , so i=1j 1

u i= 2

i=1

j 1

i-

1

i + 1= 2 1 -

1

j + 1=

2 j

j + 1.

Then j

i=1j 1u i=

j

2 jj+1

=j + 1

2, so

1

1

u 1

+2

1

u 1+

1

u 2

+3

1

u 1+

1

u 2+

1

u 3

+ +100

1

u 1+

1

u 2+ +

1

u 100

=1

2i=1

100 H j + 1L.

Yet, i=1

100 H j + 1L =i=2

101

j =H2 + 101L100

2= 5150 so the value of

1

1

u 1

+2

1

u 1+

1

u 2

+3

1

u 1+

1

u 2+

1

u 3

+ +100

1

u 1+

1

u 2+ +

1

u 100

is 2575 .

13 Answer: k = 501 .

As 2010 ! = 122010 and as 10 = 25, so it is enough to count how many 5s in 2010 ! .We have 2010 = 5402 so there are 402 multiples of 5 in the set 81; 2; ; 2010< and then at least 402 factors 5s in theproduct 2010 ! . Identically, 2010 = 5280 + 10 so there are 80 multiples of 52 in the set 81; 2; ; 2010 1 and 1

log aHbL +1

log bHaL = 1229 , find the value of 1

log a bHbL -1

log a bHaL .As a > b > 1,

1

log a bHbL -1

log a bHaL > 0.Then

1

log a bHbL -1

log a bHaL =1

lnHbLlnHa bL

-1

lnHaLlnHa bL

=lnHa bLlnHbL -

lnHa bLlnHaL =

lnHaL + lnHbLlnHbL -

lnHaL + lnHbLlnHaL =

lnHaLlnHbL -

lnHbLlnHaL .

As lnHaLlnHbL -

lnHbLlnHaL > 0,

1

log a bHbL -1

log a bHaL =lnHaLlnHbL -

lnHbLlnHaL

2

.

Lets yet note 1

log aHbL +1

logbHaL =1

lnHbLlnHaL

+1

lnHaLlnHbL

=lnHaLlnHbL +

lnHbLlnHaL so

lnHaLlnHbL -

lnHbLlnHaL

2

=lnHaLlnHbL +

lnHbLlnHaL

2

- 4lnHaLlnHbL

lnHbLlnHaL = 1229 - 4 = 1225.

Finally 1

log a bHbL -1

log a bHaL = 1225 = 35.

6 Lfs Maths Lab, SMO Senior

If a > b > 1 and 1

log aHbL +1

log bHaL = 1229 , find the value of 1

log a bHbL -1

log a bHaL .As a > b > 1,

1

log a bHbL -1

log a bHaL > 0.Then

1

log a bHbL -1

log a bHaL =1

lnHbLlnHa bL

-1

lnHaLlnHa bL

=lnHa bLlnHbL -

lnHa bLlnHaL =

lnHaL + lnHbLlnHbL -

lnHaL + lnHbLlnHaL =

lnHaLlnHbL -

lnHbLlnHaL .

As lnHaLlnHbL -

lnHbLlnHaL > 0,