Section A
1.
x 1
a x  b x  (a  b)(  (1) i a x 1i  b i )
i 0
x 1
a x  b x  (a  b)(  a x 1i  b i )
i 0
a  b (mod c) means that a and b give the same remainder when divided
by c.
14 2009  16 2009
 (14  16)(14 2008  14 2007  16  14 2006  16 2    14  16 2007  16 2008 )
 30  (14 2008  14 2007  16  14 2006  16 2    14  16 2007  16 2008 )
14 2008  14 2007  16  14 2006  16 2    14  16 2007  16 2008
 (1) 2008  (1) 2007  (1)  (1) 2006  (1) 2    (1)  (1) 2007  (1) 2008
 (1)  (1)  (1)    (1)  (1)
 11111
 2009
 14 (mod 15)
Let 142008 142007 16  142006 162  14 162007  162008  15k  14
14 2009  16 2009
 30  (15k  14)
 450k  420
 195 (mod 225)
 the remainder is 195
Alternative Method
In questions of number theory, sometimes the techniques and skills
involved may be too difficult. Finding permutation may be slow, but it
increases your chances of success if you are patient enough to find the
pattern.
Find the permutation when 14 n and 16 n are divided by 225. It can be found
that the repeating cycles consist of 30 and 15 terms respectively. By the
pattern found, it can be found that 14 2009  209 (mod 225) and
16 2009  211 (mod 225) respectively. Hence,
14 2009  16 2009  209  211  195 (mod 225)
2.
It is always easier to count something in a systematical way.
11 square(s) : 9  5  45
2 2 square(s) : 4  5  20
3 3 square(s) : 9
4 4 square(s) : 0
5  5 square(s) : 0
6  6 square(s) : 4
7  7 square(s) : 0
8 8 square(s) : 0
9  9 square(s) : 1
45  20  9  4  1  79
 there are 79 squares in the diagram.
3.
Originally, there are 1  2  3    n sweets in total.
After the students with an even number of sweets eat all their sweets,
1  3  5    k sweets are left (where k is the largest odd number not
exceeding n ).
1  3  5    k  25
1 k 2
(
)  52
2
k 9
 n  9 or 10
4.
Let the six-digit integer be ABCDEF
Let x  AB and y  CDEF
9  ABCDEF  CDEFAB
9  (10000 x  y )  100 y  x
89999 x  91 y
989 x  y
x is a two-digit integer while y is a four-digit integer.
Hence, x  10
If x  11 , y  11 989  10879 is not a four-digit integer.
Therefore, x  10 , y  9890 , and the number is 109890
5.
As B' E is the perpendicular bisector of AB , we have B' A  B' B
We also have AB'  AB by the construction of B'
Therefore AB  AB'  BB' , and ABB' is an equilateral triangle.
BAB'  60
Consider quadrilateral ABGB '
BGB'  360   GB' A  B ' AB  ABG
 360   90  60  90
 120 
B' GC  180  BGB'
 60
6.
For multiples of 9, the sum of its digit must be a multiple of 9.
For multiples of 11, the difference between sum of odd-numbered digits
and the sum of even-numbered digits must be a multiple of 11.

101
From 9 | 1010

,
n digits
9 | (1  0  1  0    1  0  1)
Therefore, there are 9k 1 ’s for some positive integer k

101
From 11 | 1010

,
n digits
11 | (1  1    1  1)  (0  0    0)
Therefore, there are 11l 1 ’s for some positive integer l
Together, there are 99m 1 ’s for some positive integer m
There are at least 99 1 ’s
 the smallest n is 197
7.
It is easy to see that 125  53 | 65 2009
65 2009  12009  1 (mod 8)
 65 2009  0 (mod 125 )
Solving  2009
 65
 1 (mod 8)
By observing 8 numbers, 65 2009  625 (mod 1000 )
 the last three digits of 652009 is 625
8.
Please remember to fill in the unit.
Assume a , b , c , d , e are the time needed to fill up the pool by opening pipe
A , B , C , D , E only, respectively.
1 1 1 1
  
a b c 4
1 1 1 1
  
b c d 5
1 1 1 1
  
c d e 6
1 1 1
1 1 1
1 1 1 1 1 1
(   )(   )(   )   
a b c
b c d
c d e
4 5 6
1 1 1 13
  
a c e 60

60
hours is needed.
13
9.
It is possible that a problem has no solution. Don’t hesitate if you find
this result and are sure all your steps are correct.
If x  1 ,
( x  1) 2  0
x2  2x  1  0
2x 1  x2
2x 1  x (x 
1
 0)
2
x@ x  x
( x @ x) @ x  ( x @ x)  x  1
 x  x 1
 x@ x
x
[( x @ x) @ x] @ x  x
Hence, there is no solution other than x  1
10.
Denote right, downwards and diagonally downwards to the right by R, D and G
respectively.
If the King does not move diagonally, it must move 3R and 3D in some
permutation.
Number of possible routes

6!
3!3!
 20
If the King moves diagonally once, it must move 1G, 2R and 2D in some
permutation.
Number of possible routes

5!
1! 2 ! 2 !
 30
Similarly, total number of possible routes is :
6!
5!
4!
3!



3 ! 3 ! 1 ! 2 ! 2 ! 2 !1 !1 ! 3 !
 20  30  12  1
 63
Section B
1.
Conditional Probability =
possible
(outcomes)
given
P(Draw a red ball and lie)
1 1

3 5

2 4 1 1
  
3 5 3 5
1
 15
8 1

15 15
1
 15
9
15
1

9
Alternative Method: List out all outcomes
2.
Note that x2  0 or 1 (mod 3)
To have (2 p 2  3) 2  ( p  2) 2  224  2  1 (mod 3)
We must have (2 p 2  3) 2  0 (mod 3) and ( p  2) 2  1 (mod 3)
From (2 p 2  3) 2  0 (mod 3)
2 p 2  3  0 (mod 3)
2 p 2  0 (mod 3)
p 2  0 (mod 3)
p  0 (mod 3)
As p is a prime, p  3
Always check your answer if the problem requires you to solve an
equation. Sometimes you may forget to reject some cases.
Checking : (2(3) 2  3) 2  (3  2) 2  15 2  12  224
Hence, p  3 is the only solution.
3.
Only square numbers can have odd number of factors.
Assume ai  x 2
If x  1 , ai  1 only has 1 positive divisor.
If x  1 , ai has 1 , x , x 2 as factors.
Hence, x does not have any other factors except 1 and x .
x is a prime.
Each ai is the square of a prime, so a10 is the square of the tenth prime, i.e.
a10  29 2  841
4.
To be the sides of a triangle, the lengths must satisfy triangle inequality.
The three lengths are x , y and 100  x  y respectively.
x  y  100  x  y


 y  (100  x  y )  x
 x  (100  x  y )  y

 x  y  50

  50  x
 50  y

When x  1 , y has 0 choices.
When x  2 , y has 1 choice.
When x  3 , y has 2 choices.

When x  49 , y has 48 choices
However, we should discard the case x  y , where there are 24 choices.
P( x , y and 100  x  y can form a triangle)
1  2    48  24
100  99
(1  48)  48
 24
2

9900
1152

9900
32

275

5.
Cosine Law: In a Triangle ABC, AC  AB 2  BC 2  2 AB  BC cos ABC
Join the segments FE , FC
ABC  FEA (ext.   opp. int . )
 BAD ( in alt. segment)
 60
BFC  90 ( in semi  circle)
BF
cos 60
 2 BF
8
BC 
By Cosine Law,
AC 
AB 2  BC 2  2 AB  BC cos 60
1
 7 2  8 2  2(7)(8)( )
2
 57
Section C
1.
The diagram is composed of four figures.
To state clearly, region I is the moon-shaped figure. Region II is the
V-shaped figure. Region III is the convex-shaped figure, and region IV is
an equilateral triangle.
Note the followings :
 I  2  II  3  III  IV

II



III  IV




IV



1
3 2
)r
 I (  
6
2

1 2

 II  6 r 

 III  ( 1   3 )r 2

6
4

 IV  3 r 2

4
 area of circle  r 2
 III  IV
1
 area of a 60 sec tor  r 2
6
r
r  r 2  ( )2
2  3 r2

2
4
Required ratio

3  III  IV
3  I  3  II  3  III  IV
1
3 3
3 2
( 

)r
2
4
4

1
3 3 1
1
3 3
3 2
( 
  

)r
2
2
2
2
4
4
 3

3  2 3

3 2  5 3  6
9 2  12
1
1 2  k
2.
1
(1  k )k
2
2

k (k  1)
1
1
 2( 
)
k k 1

k
1
1

1 2
1 2  k
k

1 1
1 1
1
1
1  2(  )  2(  )    2( 
)
2 3
3 4
k k 1
k

1
1
1  2( 
)
2 k 1
k

k 1
1  2(
)
2(k  1)
k

k 1
1
k 1
k

2k
k 1
k 1

2
1
2
1

3
1
1
1
1

1 2
1 2 1 2  3
11 2 1
2009  1



2
2
2
1
 (2  3  4    2010 )
2
1 (2010  2)(2009 )
 
2
2
 1010527
1
3.
2009

1
1
1

1 2
1  2    2009
Solution 1
S1  12  4 2  7 2    298 2
Let
,
S 2  2 2  52  82    299 2
S3  32  6 2  9 2    300 2
 32  (12  2 2  32    100 2 )
100  101  201
 9
6
S1  S 2  S3  12  22  32    300 2

300  301  601
6
S 2  S1  (22  12 )  (52  42 )    (299 2  298 2 )
 (2  1)  (5  4)    (299  298)
 3  9  15    597
(3  597 )(100 )
2
 30000

and
( S1  S 2  S3 )  ( S3 )  ( S 2  S1 )
2
300  301  601
100  101  201
 9
 30000
6
6

2
50(301  601  3  101  201)  30000

2
50(180901  60903 )  30000

2
50(119998 )  30000

2
5999900  30000

2
5969900

2
 2984950
S1 
Solution 2
12  42  7 2    298 2


100
 (3i  2)2
i 1
100
 (9i 2  12i  4)
i 1
100
9

i 2  12
i 1
100

i 1
i4
100
1
i 1
100  101  201
(1  100 )  100
 12 
 400
6
2
 3045150  60600  400
 9
 2984950