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Section A
1.
Arithmetic Series: A series of numbers that have a common difference
between terms
Sum of an Arithmetic Series =
Number of terms =
1
(first term + last term)(number of terms)
2
1
(last term - first term) + 1
2
The series is an arithmetic series.
( First term = 12, Last term = 78, Common difference = 2 )
No. of terms 
78  12
 1  34
2
The required sum

2.
1
12  7834  1530
2
15 x  3 y 17

6 x  7 y 15
1515 x  3 y   176 x  7 y 
225 x  45 y  102 x  119 y
123x  164 y
x 4

y 3
3.
Notice that
1  2  3  1  1  2  1  3  1  1  2  3  1  61
3
3
3  6  9  1  3  2  3  3  3  1  2  3  3  63
3
3
5  10  15  1  5  2  5  3  5  1  2  3  5  65

3
3
2009  4018  6027  1  2009  2  2009  3  2009  1  2  3  2009  62009
3
Therefore,
1 2  3  3  6  9  5  10  15    2009  4018  6027  6 13  33  53    2009 3

3

Similarly,
1 4  7  3  12  21  5  20  35    2009  8036  14063  28 13  33  53    2009 3


The required answer is

281
3
   2009  14
6 13  33  53    2009 3
3
3 5
3
3
3
Alternative Method
1  2  3  3  6  9  5  10  15  7  14  21  ...  2009  4018  6027
1  4  7  3  12  21  5  20  35  7  28  49  ...  2009  8036  14063
=
(1  2  3)(1  3  5  ...  2009)
(1  4  7)(1  3  5  ...  2009)
=
1 2  3
1 4  7
=
3
14
4.
Consider the powers of 3:
3, 9, 27, 81, 243, 729, 2187, 6561, …
The unit digits are:
3, 9, 7, 1, 3, 9, 7, 1, …
The sequence repeats with a period of 4.
∵ 167  41 4  3
∴ Unit digit of 583167
= Unit digit of 33
=7
5.
Triangle Inequality: The sum of lengths of two sides of a triangle must be
greater than the length of the third side.
Let x be the length of the third side. By the triangle inequality,
78  x
15  x
8  x  7 => x  1 => 1  x  15
x7 8
x 1
Since x is an integer, x = 2, 3, 4, …, 13 or 14
The sum 
1
2  1413  104 (Sum of Arithmetic Sequence)
2
6.
If the product of a group of numbers is divisible by an integer n, then at
least one of the numbers in the group is a multiple of n.
Among n3–1, n3 and n3+1, at least one is divisible by 5.
By observation, n = 4
7.
If Mr. Fu does not buy the 2.5L bottle, he needs 7 bottles of 1L, cost = $6.5×7 =
$45.5
If Mr. Fu buy 1 bottle of 2.5L, he needs 4 bottles of 1L, cost = $15 + $6.5×4 =
$41
If Mr. Fu buy 2 bottles of 2.5L, he needs 2 bottles of 1L, cost = $15×2 + $6.5×2
= $43
If Mr. Fu buy 3 or more bottles of 2.5L, cost ≥ $15×3 = $45
∴ Mr. Fu must pay at least $41.
8.
possible
(number of outcomes)
total
Probability =
Total number of outcomes of rolling two dice  6  6  36
Outcomes with sum = 8: 2,6, 3,5, 4,4, 5,3, 6,2 (5 outcomes)
The probability required 
5
36
1
2
3
4
5
6
1
2
3
4
5
6
7
2
3
4
5
6
7
8
3
4
5
6
7
8
9
4
5
6
7
8
9
10
5
6
7
8
9
10
11
6
7
8
9
10
11
12
9.
Let A, B, C, D be the vertices of the quadrilateral, O be the intersection of the
diagonals.
Area of ABCD = Area of △ABC + Area of △ADC
=
1
 AC OD   1  AC OB 
2
2
1
= ( AC )(OD  OB )
2
1
1
= ( AC )( BD )  68  24
2
2
10.
Cow Eat Grass Problem: Common Question in Lower Level Competitions
Assumed cows eat grass and grass grow at constant rates.
Let the original amount of grass on the farm be 1 unit.
Let x be the units of grass eaten by a cow per week,
y be the units of grass grown per week.
1
........ (1)
32
1
40 x  y  ........ (2)
8
28 x  y 
(2) – (1), 12 x 
Sub x 
1 1
3
1


, x
8 32 32
128
1
3
1
 1 
into (1), 28
, y
 y 
128
16
32
 128 
3  1 
∴ The maximum no. of cows can be kept  y  x     
  24
 16   128 
Section B
1.
Definition: Function y=f(x)
y is a function of x if and only if each value of x corresponds to only one
value of y.
f 0  2
a0  b0  c  2
3
c2
f 3  1
a3  b3  2  1
27a  3b  1
3
f  3  a 3  b 3  2
 2 7a  3b   2
  1  2
3
3
2.
x  1# x  3  x  3# x  1
 x  1x  3  x  1  x  3  4
 x 2  4 x  3  2 x  4  4
 x  2x  3
2
x
2

 2x  3
x
2
x 2  2 x 3


4
 2x  3  2
x 2  2x  1  0
x  12  0
x  1
3.
In Fill in the Blanks Questions, look for same digits in the equation, and
note that ‘9’ can usually be found at the leftmost digits in addition.
Suppose I ≤ 7, TIME + IS + UP ≤ T799 + 99 + 99 < T999
This is impossible since the thousands-digit of the sum is S.
Suppose I = 8,
The carry from the tens is at most 2, so the hundreds-digit of the sum T
must be 0, but this does not satisfy the requirement of the question. (T is a
digit from 1 to 9)
Therefore, I = 9. It follows that T = 1 and S = 2.
Consider the unit digit,
∵ ‘E + S + P = P’ ∴ E + S = 10, E = 10 – 2 = 8
Consider the tens-digit,
M + I + U + 1 (carry from the ones) = O + 20 (carry to the hundreds)
M + U = O + 10
Since 1, 2, 7, 8 are already used, M, U, O = 3, 4, 5, 6 or 7.
The only solutions are (M = 6, U = 7, O = 3) or (M = 7, U = 6, O = 3)
In both cases, O = 3
4.
Note that in Secondary School, we express answers in terms of 
instead of taking it as a specific numerical value such as 3.14.
Area not covered
= Area of the square – Area of the circle
1
2
 1    
2
2
1/2
1
 1 
4
1
5.
n   n  1  n  2  2009  3n  3  2009  3n  2006  n  668 2
3
For n = 1, 2, 3, … , 668 , 3n  3 can be represented as the sum of three
consecutive positive integers.
Therefore, there are 668 such numbers in total.
Section C
1.
In figure (I), it is easy to see that the areas of regions A, B, C, D are the same.
For convenience, we can move region A to region C, and move region B to
region D, without changing the area of the shaded region. (as shown as Figure
(II))
Without loss of generality, let r be the radius of the quarter-circle.
The required ratio 
1
r 2  1 r 2   2
Area of the shaded region
 4 1 22 
Area of the quarter - circle

4 r
2.)
Let x, y, z be the no. of roosters, hens and chicks respectively.
x  y  z  100......(1)
5 x  3 y  0.2 z  100......(2)
(2) × 5 – (1), 24 x  14 y  400 , y 
400  24 x 200  12 x 450  3x 


14
7
7
∵ x, y are integers. ∴ 50 - 3x is divisible by 7.
(x, y) = (5, 20) or (12, 8)
Sub (x, y) = (5, 20) into (1), z  100  5  20  75
Solution 1: Derek has 5 roosters, 20 hens and 75 chicks.
Sub (x, y) = (12, 8) into (1), z  100  12  8  80
Solution 2: Derek has 12 roosters, 8 hens and 80 chicks.