1 Let 2015... 2521 aaa , where 2521

,...,, aaa are positive integers. If d is the

greatest common divisor of 2521,...,, aaa

, find the largest possible value of d.

Ans: Since 311352015 and daaa 25...2015 2521 , we must have

8025

2015

d

. It follows that the largest possible value of d is 65.

This could be attained by 65... 2921 aaa and

13030 a .

2 Find the number of integer pairs ba, which satisfy 201422 baba ?

Ans: It is easy to see that ba, must be both even. Otherwise, 22 baba is odd.

However, if ba, are both even, 22 baba is a multiple of 4. This is impossible.

Hence, there is no solution, i.e., 0 pairs.

3 Given two fair dice, one with the numbers 4,3,3,2,2,1 on its faces and the other

with the numbers 8,6,5,4,3,1 on its faces, one throws the two dice together and

obtains two numbers. If the probability that the sum of these two numbers is 5

equals %x , find the largest integer not exceeding x.

Ans: We denote the numbers on the dice 4,3,3,2,2,1 2121 and 8,6,5,4,3,1 respectively.

Out of the 36 possible outcomes, 1,4,3,2,3,2,4,1 21 give a sum of 5.

Hence, the probability is 9

1

, about 11%.

4 Given a function f such that xfxf 20152015 for any real number x, if

f has exactly four distinct real roots dcba ,,, , find dcba .

Ans: Since for any real number x, xfxf 20152015 , we conclude that the

graph of f is symmetric about the line 2015x .

Hence, 42015dcba 8060, because the roots are symmetric about the

line 2015x .

5 Tom and Jerry each has pocket money in S$1 coins. If Tom gives Jerry n dollars,

then Jerrys money is twice Toms money. If Jerry gives Tom S$2, then Toms

money is n times Jerrys money. Find the sum of all possible value of n.

Ans: Suppose Tom and Jerry has yx, dollars respectively.

We have

22

2

ynx

nxny.

It is easy to see that 22 nnyx and hence, 232 nnyny .

Now 12

47

n

ny , i.e.,

12

157

12

8142

nn

ny .

This is only possible when 5 ,3 ,112 n or 15.

One may check that 3 ,2 ,1n and 8 satisfy the condition, i.e., the sum is 14.

6 Let cba ,, be real numbers such that

01512

010222

2

acbcb

abca.

Find the product of the smallest and largest possible value of a.

Ans: We have 1022 aabc .

Hence, 10215122222 aaacbcb , i.e.,

222 52510 aaacb .

It follows that 10245 22 aaa , or 0562 aa .

Hence, 51 a and the sum is 6.

CB

A

PQ

7 Find the largest positive integer n satisfying: there exists a unique positive integer

k such that 13

7

15

8

kn

n.

Ans: We have 7

13

8

15

n

kn, i.e.,

7

6

8

7

n

k

One may write it as 112

96

112

98

n

k.

If 112n , 97k is the unique solution.

If 112n , one sees that n

n

n

k

n

n

112

96

112

112

112

98 has more than one solution,

because there exist at least two multiples of 112 between n96 and n98 .

In conclusion, the largest possible value of n is 112.

8 In ABC , 35AB , 56BC and 49AC . QP, are points on ACAB,

respectively such that the circumcircle of APQ is tangent to BC. Refer to the

diagram below on the right.

Find the smallest possible length of PQ.

Ans: Let R be the circumradius of APQ .

Notice that ARPQ sin2 where A is a constant.

Hence, PQ attains the smallest value when R is the smallest.

Since the circle touches BC, the diameter is at least the height from A to BC.

It is easy to see that the height is ABC

AACAB

sin

8

735sin.

Hence, the minimal value of PQ is AA 22 cos18

735sin

8

735

2222

21

8

735

ACAB

BCACAB

2222

752

8751

8

735

49

48

8

735

7

11

8

7352

30.

Note: One should avoid tedious calculation!

DO

A

BC

E

F

G

D

O

A

BC

9 Find the smallest prime number p such that ppp 96 23 has exactly 30 positive

factors.

Ans: ppp 96 23 23 pp . Clearly, 2p or 3 does not satisfy the condition.

Now p and 3p are coprime. It follows that 23p has 15 positive factors.

Notice that 3p is even. We must have Mp 23 2 , where is even and

15|1 .

If 14 , 1M and 142 23 p . We have 131327 p .

If 4 , M has 3 positive factors. We must have 2qM .

When 3q , 242 323 p , i.e., 15p .

When 5q , 242 523 p , i.e., 23p .

If 2 , M has 5 positive factors. We must have 4qM .

When 3q , 422 523 p , i.e., 53p is the smallest.

In conclusion, 23 is the smallest such prime number.

10 Refer to the diagram on the right. AB touches O at A.

C is a point inside O and BC intersects O at D.

Given 6AB , 3CDBD , 2OC and the area of the circle is k , find k.

Ans: Refer to the diagram below.

We have 2ABBEBD by Tangent Secant Theorem.

Hence, 12BE and 6CE .

By Intersecting Chords Theorem, CGCFCECD 18

Notice that 42 rOCrOCrCGCF , where r denotes the radius of the

circle. It follows that 2r 22.

11 Let 201521 ,...,, xxx be positive real numbers. Find the smallest possible value of

201521201421

2015

321

4

21

3

1

21

...

4

......

xxxxxx

x

xxx

x

xx

x

x

xx .

Ans: Applying AM-GM to the last two terms gives

201521201421

2015

321

4

21

3

1

21

...

4

......

xxxxxx

x

xxx

x

xx

x

x

xx

2201421201321

2014

321

4

21

3

1

21

...

42

......

xxxxxx

x

xxx

x

xx

x

x

xx

201421201321

2014

321

4

21

3

1

21

...

4

......

xxxxxx

x

xxx

x

xx

x

x

xx

Now apply the argument repeatedly. We have

201521201421

2015

321

4

21

3

1

21

...

4

......

xxxxxx

x

xxx

x

xx

x

x

xx

1

1

4

xx 4.

This could be achieved when 2... 201521 xxx .

12 For any positive integer n, we define na as the smallest positive integer such that

! na is a multiple of n. If 5

2

n

an , find n.

Ans: Since nan5

2 , n must be a multiple of 5.

Let kn 5 . Notice that if 6k , !|5 kk , which implies kan .

This is impossible because nn

k5

2

5 .

Hence, 4 ,3 ,2 ,1k or 5.

Notice that 52015105 aaaa and 1025 a , i.e., n 25.

QD

C

A

P O

B

EQ

D

C

A

PO

B

13 How many sequences 921 ,...,, aaa could be formed using 9,...,2,1 without

repetition such that each term in the sequence is either greater than every term in

front of it, or smaller than every term in front of it? For example, 543678921 is

one such sequence.

Ans: We consider the general case for naaa ,...,, 21 using n,...,2,1 .

Let nI denote the number of sequences satisfying the condition.

Suppose nai . We must have 1na because it must be smaller than every term

in front of it. Similarly, 21 na , 32 na and inai 1 .

For 121 ,...,, iaaa , there are exactly 1iI way to arrange the remaining numbers.

Hence, 121 ...1 nn IIII .

Notice that 11221 2...1 nnnn IIIIII .

Since 22 I , we have 12 nnI and in particular, 9I 256.

14 Refer to the diagram below on the right. Given that the radius of O is 12, P is a

point outside O and 24OP . A line passing through P intersects O at BA, .

Let DC , be the point symmetric to BA, about the line OP. If BCAD, intersect

at Q, find OQ.

Ans: Notice that BOEBODBAQ 2

1.

Hence, QOBA ,,, are concyclic.

We have PBPAPQPO by Tangent Secant Theorem.

Now PBPA is the power of P with respect to O and one may show (by

Tangent Secant Theorem) that 22222 1231224 OEOPPBPA .

It follows that 1824

123 2

OP

PBPAPQ .

Hence, PQOPOQ 6.

15 Movie tickets are sold at S$5 each. 10 people queue up to purchase the tickets,

where 5 of them each hold a five-dollar note, and the other 5 each hold a ten-

dollar note. However, the receptionist did not bring any cash. These 10 people

queue up randomly, and the receptionist will be complained if he fails to give

change to anyone. If the probability that the receptionist will not be complained is

%x , find the largest integer not exceeding x.

Ans: There are

2525

10

ways to receive 5 five-dollar notes and 5 ten-dollar notes

sequentially.

Note that this is equivalent to the number of ways travelling from A to B in 10

unit length in the following diagram. One may perceive travelling towards right

for one unit as receiving a five-dollar note, while travelling downwards for one

unit means receiving a ten-dollar note.

If the receptionist is not complained, it means at any stage, he would not receive

more ten-dollar notes than five-dollar notes. This is equivalent to travelling from

A to B while staying in the upper triangular region and not crossing the main

diagonal. One may easily see that there are 42 ways (using addition).

Hence, the probability is 6

1

252

42

, about 16%.