Division
S4
Singapore and Asian
Schools Math Olympiad
2015
Full Name:
Index Number:
Class:
School:
SASMO 2015 Secondary 4 Contest
INSTRUCTIONS
1. Please DO NOT OPEN the contest booklet until the Proctor has given permission to
start.
2. TIME: 1 hour 30 minutes.
3. Attempt all 25 questions.
Questions 1 to 15 score 2 points each, no points are deducted for unanswered
question and 1 point is deducted for wrong answer.
Questions 16 to 25 score 4 points each. No points are deducted for unanswered or
wrong answers.
4. Shade your answers neatly using a pencil in the answer sheet.
5. PROCTORING: No one may help any student in any way during the contest.
6. No electronic devices capable of storing and displaying visual information is
allowed during the course of the exam.
7. Strictly No Calculators are allowed into the exam.
8. All students must fill and shade in their Name, Index number, Class and School in
the answer sheet and contest booklet.
9. MINIMUM TIME: Students must stay in the exam hall at least 1h 15 min.
10. Students must show detailed working and transfer answers to the answer sheet.
11. No exam papers and written notes can be taken out by any contestant.
10
SASMO 2015, Secondary 4 Contest
SASMO 2015 Secondary 4
Starting Score = 15 marks (to avoid negative marks); Max Possible Score = 85
marks
Section A (Correct answer = 2 marks; no answer = 0; incorrect answer =
minus 1 mark)
1.
The diagram shows two overlapping squares. The length of the bigger square
is 14 cm and the length of the smaller square is 7 cm. Find the difference
between the area of the two unshaded regions X and Y.
X
Y
(a)
(b)
(c)
(d)
(e)
137 cm2
147 cm2
157 cm2
167 cm2
None of the above
________________________________________________________________
2.
In a class of 40 students, 7 study both Physics and Chemistry, 16 study
Physics and 14 study Chemistry. How many students do not study either
Physics or Chemistry?
(a)
(b)
(c)
(d)
(e)
3
7
10
17
None of the above
12
SASMO 2015, Secondary 4 Contest
3.
Find the range of values of 𝑘 if the curve 𝑦 = 𝑘𝑥 2  2 𝑥 + (2 𝑘 – 1) lies
completely above the 𝑥-axis.
1
<𝑘<1
(a)

(b)
𝑘<
(c)
(d)
(e)
0<𝑘<1
𝑘>1
None of the above
2
1
2
or 𝑘 > 1
________________________________________________________________
4.
A number gives a remainder of 9 when divided by 10. Another number gives
a remainder of 8 when divided by 10. The sum of these two numbers is
multiplied by 12 to give the third number. What is the remainder when this
third number is divided by 10?
(a)
(b)
(c)
(d)
(e)
4
7
8
9
None of the above
13
SASMO 2015, Secondary 4 Contest
5.
In the figure below, the ratio of the trapezium ABCD to the area of the
1
triangle DEF to the area of parallelogram GHJK is 4 : 2 : 3. Given that 3 of the
area of ∆DEF is shaded, find the ratio of the area of the shaded region to the
total area of the unshaded regions of the figure.
B
C
F
G
A
H
D
E
J
K
(a)
(b)
(c)
(d)
(e)
2 : 11
2 : 17
2 : 21
2 : 27
None of the above
________________________________________________________________
6.
Which of the following statement(s) is or are correct?
Statement A: A cubic equation can have 3 real and distinct roots.
Statement B: A cubic equation can have 2 real roots.
Statement C: A cubic equation can have 1 real root and 2 non-real roots
(a)
(b)
(c)
(d)
(e)
All the three statements are correct.
Only Statements A and B are correct.
Only Statements A and C are correct.
Only Statement A is correct.
None of the above
14
SASMO 2015, Secondary 4 Contest
7.
A big cube is made up of 125 small cubes. All the faces of the big cube are
then painted. How many of the small cubes have no painted face?
(a)
(b)
(c)
(d)
(e)
1
8
27
64
None of the above
________________________________________________________________
8.
In ABC, AB = 14 cm, BC = 10 cm and AC = 7 cm. Find the value of
(a)
(b)
(c)
(d)
(e)
0.5
1
2
Cannot be found
None of the above
15
sin 𝐵
sin 𝐶
.
SASMO 2015, Secondary 4 Contest
9.
All the match sticks in the diagram are identical. Find the total number of
squares in the diagram?
(a)
(b)
(c)
(d)
(e)
6
7
8
9
None of the above
________________________________________________________________
10.
Given that 𝑥𝑦𝑧 = 2015, and 𝑥 , 𝑦 and 𝑧 are positive integers, how many
possible triples (𝑥, 𝑦, 𝑧) are there?
(a)
(b)
(c)
(d)
(e)
5
15
27
2015
None of the above
16
SASMO 2015, Secondary 4 Contest
11.
𝑥
Given that 8  𝑥  12 and 4  𝑦  2, find the least possible value of 𝑦.
(a)
(b)
(c)
(d)
(e)
4
2
3
6
12
________________________________________________________________
12.
How many four-digit numbers of the form X56Y are divisible by 24?
(a)
(b)
(c)
(d)
(e)
1
3
4
6
8
17
SASMO 2015, Secondary 4 Contest
13.
A rectangular floor of 1540 cm by 1440 cm is to be covered completely by
identical square tiles. What is the least possible number of square tiles?
(a)
(b)
(c)
(d)
(e)
616
5544
22 176
88 704
None of the above
________________________________________________________________
14.
Johnny has 35 toys. He divides them into 5 piles so that each pile has a
different number of toys. Find the smallest possible number of toys in the
biggest pile.
(a)
(b)
(c)
(d)
(e)
7
8
9
10
None of the above
18
SASMO 2015, Secondary 4 Contest
15.
The diagram shows a circle with two chords AB and CD intersecting at E.
Given that AE = 12 cm, BE = 3 cm and CE = 9 cm, find the length of DE.
C
9
3
12
E
B
D
A
(a)
(b)
(c)
(d)
(e)
2 cm
3 cm
4 cm
6 cm
None of the above
________________________________________________________________
Section B (Correct answer = 4 marks; incorrect or no answer = 0)
16.
A man buys 30 metres of fence to build a rectangular garden at the back of
his house. He uses the wall XY at the back of his house as one side of the
garden ABCD as shown in the diagram below. Find the largest possible area
of the garden.
X
wall
A
D
B
C
19
Y
SASMO 2015, Secondary 4 Contest
17.
In a school hall,
7
31
of the chairs are arranged in rows of 5, and
11
31
of the chairs
are arranged in rows of 13. The rest of the chairs are stacked up. If there are
less than 4000 chairs in the hall, find the total number of chairs in the hall.
________________________________________________________________
18.
Polite numbers are numbers that can be expressed as the sum of two or more
consecutive positive integers, e.g.
5 = 2 + 3;
9 = 2 + 3 + 4 = 4 + 5.
The degree of politeness of a number is the number of ways a number can be
expressed as the sum of two or more consecutive positive integers, e.g. the
degree of politeness of 2, 5 and 9 is 0, 1 and 2 respectively.
Find the smallest number with a degree of politeness of 3.
20
SASMO 2015, Secondary 4 Contest
19.
Find the value of √12 + √12 + √12+. . .
________________________________________________________________
20.
The figure shows a circle with 4 points on its circumference. Each point is
joined to every other point by a line (called a chord). The chords divide the
circle into 8 regions.
Find the maximum number of regions formed by the chords for a circle with 7
points.
21
SASMO 2015, Secondary 4 Contest
21.
Find the values of 𝑘 for which the equation 𝑘𝑥 2  2015𝑥 + (𝑘 – 2015) = 0 has
one positive and one negative root.
________________________________________________________________
22.
A circle and a triangle are drawn on a rectangular sheet of paper. What is the
biggest number of regions that can be formed on the paper?
22
SASMO 2015, Secondary 4 Contest
23.


3

4
2
2
Find the sum of the coefficients in the expansion of 6 x  5 x  2 3  2 x  x .
________________________________________________________________
24.
Albert and Bernard just become friends with Cheryl, and they want to know
when her birthday is. Cheryl gives them a list of 10 possible dates.
May 15
June 17
July 14
August 14
May 16
June 18
July 16
August 15
May 19
August 17
Cheryl then tells Albert and Bernard separately the month and the day of her
birthday respectively.
Albert:
Bernard:
Albert:
I don’t know when Cheryl’s birthday is, but I know that Bernard
does not know too.
At first I don’t know when Cheryl’s birthday is, but I know now.
Then I also know when Cheryl’s birthday is.
So when is Cheryl’s birthday?
23
SASMO 2015, Secondary 4 Contest
25.
Find the last six digits of 20152015.
END OF PAPER
24
SASMO 2015, Secondary 4 Contest
SASMO 2015 Secondary 4 Solution
Section A
1.
The diagram shows two overlapping squares. The length of the bigger square is 14 cm
and the length of the smaller square is 7 cm. Find the difference between the area of
the two unshaded regions X and Y.
X
Y
137 cm2
147 cm2 [Ans]
157 cm2
167 cm2
None of the above
(a)
(b)
(c)
(d)
(e)
Solution
Let the area of the overlapping region be A.
Area of X  Area of Y = (Area of big square  A)  (Area of small square  A)
= Area of big square  A  Area of small square + A
= Area of big square  Area of small square
= 142  72
= 196  49
= 147 cm2
2.
In a class of 40 students, 7 study both Physics and Chemistry, 16 study Physics and 14
study Chemistry. How many students do not study either Physics or Chemistry?
(a)
(b)
(c)
(d)
(e)
3
7
10
17 [Ans: 40  9  7  7]
None of the above
Solution
Fill in the following Venn diagram in this order:




No. of students who study both Physics (P) and Chemistry (C) = 7
No. of students who study Physics only = 16  7 = 9
No. of students who study Chemistry only = 14  7 = 7
No. of students who do not study either Physics or Chemistry
= 40  9  7  7 = 17
25
SASMO 2015, Secondary 4 Contest

P
C
9
7
7
10
 no. of students who do not study either Physics or Chemistry = 17
3.
Find the range of values of k if the curve y = kx2  2x + (2k – 1) lies completely above
the x-axis.
1
(a)
2<k<1
(b)
(c)
(d)
(e)
k <  2 or k > 1
0<k<1
k > 1 [Ans]
None of the above
1
Solution
If the curve y = kx2  2x + (2k – 1) lies completely above the x-axis, then b2 – 4ac < 0.
4 – 4k(2k – 1) < 0
1 – k(2k – 1) < 0
1 – 2k2 + k < 0
2k2 – k – 1 > 0
(2k + 1)(k – 1) > 0
1
k <  2 or k > 1
But the coefficient of x2 must be positive for y = kx2  2x + (2k – 1) to lie completely
above the x-axis.
 k > 1.
4.
A number gives a remainder of 9 when divided by 10. Another number gives a
remainder of 8 when divided by 10. The sum of these two numbers is multiplied by
12 to give the third number. What is the remainder when this third number is divided
by 10?
(a)
(b)
(c)
(d)
(e)
4 [Ans: 9 + 8 = 17  7  2 = 14]
7
8
9
None of the above
26
SASMO 2015, Secondary 4 Contest
Solution
Method 1
1st number = 10 + 10 + 10 + … + 10 + 9 (it does not matter how many 10s are there)
2nd number = 10 + 10 + 10 + … + 10 + 8 (it does not matter how many 10s are there)
Sum of first two numbers = 10 + 10 + 10 + … + 10 + 17
= 10 + 10 + 10 + … + 10 + 10 + 7
rd
3 number = 12  sum of first two numbers
= 10 + 10 + 10 + … + 10 + 84 (it does not matter how many 10s are there)
= 10 + 10 + 10 + … + 10 + 10 + 10 + 10 + … + 10 + 4
 when the 3rd number is divided by 10, the remainder is 4.
Method 2
Let the first number be 10x + 9 and the second number be 10y + 8, where x and y are
whole numbers.
Then the sum of the first two numbers is (10x + 9) + (10y + 8) = 10(x + y) + 17
= 10(x + y + 1) + 7 = 10z + 7 for some whole number z.
Thus the third number is 12(10z +7) = 120z + 84 = 120z + 80 + 4 = 10(12z + 8) + 4
= 10w + 4 for some whole number w.
 when the third number is divided by 10, the remainder is 4.
5.
In the figure below, the ratio of the trapezium ABCD to the area of the triangle DEF
to the area of parallelogram GHJK is 4 : 2 : 3. Given that
1
3
of the area of DEF is
shaded, find the ratio of the area of the shaded region to the total area of the unshaded
regions of the figure.
B
C
F
G
A
D
E
J
K
(a)
(b)
(c)
(d)
(e)
H
2 : 11
2 : 17 [Ans]
2 : 21
2 : 27
None of the above
27
SASMO 2015, Secondary 4 Contest
Solution
Since the number 2, in the ratio 4 : 2 : 3, is not divisible by 3, we use the equivalent
ratio 12 : 6 : 9 because 6 is divisible by 3.
Let the area of DEF be 6 units.
1
Area of shaded region = 6 = 2 units
3
Area of trapezium ABCD = 12 units
Area of parallelogram GHJK = 9 units
Area of trapezium ABCD that is unshaded = 12  2 = 10 units
Area of parallelogram GHJK that is unshaded = 9  2 = 7 units
Total area of unshaded regions = 10 + 7 = 17 units
 ratio of area of shaded region to total area of unshaded regions = 2 : 17
6.
Which of the following statement(s) is or are correct?
Statement A: A cubic equation can have 3 real and distinct roots.
Statement B: A cubic equation can have 2 real roots.
Statement C: A cubic equation can have 1 real root and 2 non-real roots
(a)
(b)
(c)
(d)
(e)
All the three statements are correct.
Only Statements A and B are correct.
Only Statements A and C are correct. [Ans]
Only Statement A is correct.
None of the above
Solution
A cubic equation has exactly 3 roots:



3 real and distinct roots (when the curve cuts the x-axis at 3 distinct points)
2 real and equal roots, and another distinct real root (when the curve just
touches the x-axis at 1 point, which gives rise to 2 real and equal roots; and the
curve cuts the x-axis at another point)
1 real root and 2 non-real roots (when the curve cuts the x-axis at only 1 point)
 only Statements A and C are correct.
7.
A big cube is made up of 125 small cubes. All the faces of the big cube are then
painted. How many of the small cubes have no painted face?
(a)
(b)
(c)
(d)
(e)
1
8
27 [Ans: 33 = 27]
64
None of the above
28
SASMO 2015, Secondary 4 Contest
Solution
Since 125 = 5  5  5, then the dimensions of the big cube are 5 by 5 by 5.
The only small cubes that have no painted face are those that are not visible from
outside.
 there are (5  2)3 = 33 = 27 small cubes that have no painted face.
8.
In ABC, AB = 14 cm, BC = 10 cm and AC = 7 cm. Find the value of
(a)
(b)
(c)
(d)
(e)
sin 𝐵
sin 𝐶
.
0.5 [Ans]
1
2
Cannot be found
None of the above
Solution
Using Sine Rule,

9.
sin 𝐵
7
=
sin 𝐶
14
sin 𝐵
, i.e.
=
𝑏
sin 𝐵
sin 𝐶
sin 𝐶
𝑐
=
7
14
, where b = AC = 7 and c = AB = 14.
=
𝟏
𝟐
or 0.5
All the match sticks in the diagram are identical. Find the total number of squares in
the diagram?
(a)
(b)
(c)
(d)
(e)
6
7
8 [Ans]
9
None of the above
Solution
No. of 1  1 squares = 5
No. of 2  2 squares = 2
No. of 3  3 squares = 1
 total no. of squares = 8
29
SASMO 2015, Secondary 4 Contest
10.
Given that xyz = 2015, and x, y and z are positive integers, how many possible triples
(x, y, z) are there?
(a)
(b)
(c)
(d)
(e)
5
15
27 [Ans]
2015
None of the above
Solution
2015 = 5  13  31, where 5, 13 and 31 are prime numbers.
 2015 = 1  1  2015 (3 arrangements)
= 1  5  403 (3! = 6 arrangements)
= 1  13  155 (3! = 6 arrangements)
= 1  31  65 (3! = 6 arrangements)
= 5  13  31 (3! = 6 arrangements)
 there are 27 possible triples (x, y, z).
11.
𝑥
Given that 8  x  12 and 4  y  2, find the least possible value of .
𝑦
(a)
(b)
(c)
(d)
(e)
4
2
3
6 [Ans]
12
Solution
Because the values of x and y can be negative, we have to test different combinations
of the greatest and least possible values of x and y as follow:
−8
−4
−8
−2
12
−4
12
−2
=2
=4
= 3
= 6
 least possible value of
𝑥
𝑦
= 6
30
SASMO 2015, Secondary 4 Contest
12.
How many four-digit numbers of the form X56Y are divisible by 24?
(a)
(b)
(c)
(d)
(e)
1
3
4
6 [Ans: 1560, 4560, 7560, 2568, 5568, 8568]
8
Solution
If a number is divisible by 24 (= 3  8), then it is also divisible by 3 and 8, since 3 and
8 are relatively prime.
Using the divisibility test for 8, the last three digits 56Y is also divisible by 8. But
there are still 10 possibilities for Y.
If a number is divisible by 8, then the number is also divisible by any factor of 8, i.e.
X56Y is also divisible by 2 and by 4.
Since X56Y is divisible by 2, then Y must be even, i.e. Y = 0, 2, 4, 6 or 8.
Since X56Y is divisible by 4, using the divisibility test for 4, the last two digits 6Y is
also divisible by 4. Since only 62 and 66 are not divisible by 4, then Y = 0, 4 or 8.
Now we use the divisibility test for 8. Since only 564 is not divisible by 8, then Y = 0
or 8.
Using the divisibility test for 3, X + 5 + 6 + Y = X + Y + 11 is also divisible by 3.
If Y = 0, then X + Y + 11 = X + 11 is also divisible by 3, i.e. X = 1, 4 or 7.
If Y = 8, then X + Y + 11 = X + 19 is also divisible by 3, i.e. X = 2, 5 or 8.
Thus four-digit numbers of the form X56Y that are divisible by 24 are: 1560, 4560,
7560, 2568, 5568, 8568.
 no. of four-digit numbers of the form X56Y that are divisible by 24 = 6
13.
A rectangular floor of 1540 cm by 1440 cm is to be covered completely by identical
square tiles. What is the least possible number of square tiles?
(a)
(b)
(c)
(d)
(e)
616
5544 [Ans]
22 176
88 704
None of the above
Solution
1540 = 22 5  7  11
1440 = 25  32  5
Possible lengths of the square tiles must be factors of both 1540 cm and 1440 cm.
Greatest possible length of the square tiles = HCF of 1540 cm and 1440 cm
= 22  5
= 20 cm
The least possible number of square tiles occurs when the square tiles are the largest.
 least possible number of square tiles =
31
1540×1440
20×20
SASMO 2015, Secondary 4 Contest
=
22 ×5×7×11×25 ×32 ×5
22 ×5×22 ×5
= 7  11  23  32
= 5544
14.
Johnny has 35 toys. He divides them into 5 piles so that each pile has a different
number of toys. Find the smallest possible number of toys in the biggest pile.
(a)
(b)
(c)
(d)
(e)
7
8
9 [Ans]
10
None of the above
Solution
For each pile to have a different number of toys, and the biggest pile to have the
smallest possible number of toys, put 1 toy in the 1st pile, 2 toys in the 2nd pile, 3
toys in the 3rd pile, 4 toys in the 4th pile and 5 toys in the 5th pile. So the biggest pile
is the 5th pile, but there are only 1 + 2 + 3 + 4 + 5 = 15 toys.
The 16th toy will have to go to the 5th pile so that each pile will have a different
number of toys. The 17th toy cannot go to the 5th pile because we want to find the
smallest possible number of toys in the biggest pile, so the 17th toy will have to go
to the 4th pile. Similarly, the 18th, 19th and 20th toys will go to the 3rd, 2nd and 1st
piles respectively.
The 21st toy will then go to the 5th pile again, and the 22nd, 23rd, 24th and 25th toys will
go to the 4th, 3rd, 2nd and 1st piles respectively.
Similarly, the 26th to 30th toys will go to the 5th to 1st piles respectively.
Similarly, the 31st to 35th toys will also go to the 5th to 1st piles respectively.
 the largest pile (which is the 5th pile) will contain 5 + 1 + 1 + 1 + 1 = 9 toys.
32
SASMO 2015, Secondary 4 Contest
15.
The diagram shows a circle with two chords AB and CD intersecting at E. Given that
AE = 12 cm, BE = 3 cm and CE = 9 cm, find the length of DE.
C
9
3
12
E
B
D
A
(a)
(b)
(c)
(d)
(e)
2 cm
3 cm
4 cm [Ans]
6 cm
None of the above
Solution
Method 1
DBE and ACE are similar (angles in same segment and vertically opposite angles).

𝐷𝐸
𝐵𝐸
=
𝐴𝐸
𝐶𝐸
, i.e.
𝐷𝐸
3
=
12
9
, so DE = 4 cm.
Method 2
If you know about the intersecting chords theorem, i.e. DE  CE = AE  BE, then
DE  9 = 12  3, so DE = 4 cm.
Note: To prove the intersecting chords theorem, use similar triangles as in Method 1.
33
SASMO 2015, Secondary 4 Contest
Section B
16.
A man buys 30 metres of fence to build a rectangular garden at the back of his house.
He uses the wall XY at the back of his house as one side of the garden ABCD as
shown in the diagram below. Find the largest possible area of the garden.
X
wall
A
D
B
C
Y
Solution
Let BC = x m.
30  x
 30 x  x 2  2
 30  x 
 m .
Then AB =
m and area of garden ABCD = 
 x = 
2
2
 2 


 30  x 
Let y  
 x.
 2 
This is a quadratic function and its graph is an inverted-U parabola which cuts the xaxis at x = 0 and x = 30 as shown in the diagram below.
y
0
15
30 x
0  30
= 15, i.e. BC = 15 m.
2
 AB = 7.5 m and area of garden = 15  7.5 = 112.5 m2
By symmetry, the maximum point occurs when x =
34
SASMO 2015, Secondary 4 Contest
17.
7
11
In a school hall, of the chairs are arranged in rows of 5, and of the chairs are
31
31
arranged in rows of 13. The rest of the chairs are stacked up. If there are less than
4000 chairs in the hall, find the total number of chairs in the hall.
Solution
Let the total number of chairs in the hall be n.
Since
7
31
 n chairs are arranged in rows of 5, and the numerator 7 is not divisible by 5,
then n must be divisible by 5.
11
Since  n chairs are arranged in rows of 13, and the numerator 11 is not divisible
31
by 13, then n must also be divisible by 13.
7
11
Since  n and  n are whole numbers, then n must also be divisible by 31.
31
31
Since 5, 13 and 31 have no common factors other than 1, then n must be a multiple of
5  13  31 = 2015.
Since 2  2015 = 4030 is already more than 4000, then n = 2015.
 total no. of chairs in the hall = 2015
18.
Polite numbers are numbers that can be expressed as the sum of two or more
consecutive positive integers, e.g.
5 = 2 + 3;
9 = 2 + 3 + 4 = 4 + 5.
The degree of politeness of a number is the number of ways a number can be
expressed as the sum of two or more consecutive positive integers, e.g. the degree of
politeness of 2, 5 and 9 is 0, 1 and 2 respectively.
Find the smallest number with a degree of politeness of 3.
Solution
Method 1
1
2
3=1+2
4
5=2+3
6=1+2+3
7=3+4
8
9=4+5=2+3+4
10 = 1 + 2 + 3 + 4
11 = 5 + 6
12 = 3 + 4 + 5
13 = 6 + 7
14 = 2 + 3 + 4 + 5
15 = 7 + 8 = 4 + 5 + 6 = 1 + 2 + 3 + 4 + 5
 the smallest number with a degree of politeness of 3 is 15.
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SASMO 2015, Secondary 4 Contest
Method 2
1+2=3
2+3=5
3+4=7
⋮
1+2+3=6
2+3+4=9
3 + 4 + 5 = 12
⋮
1 + 2 + 3 + 4 = 10
2 + 3 + 4 + 5 = 14
3 + 4 + 5 + 6 = 18
⋮
1 + 2 + 3 + 4 + 5 = 15
2 + 3 + 4 + 5 + 6 = 20
3 + 4 + 5 + 6 + 7 = 25
Each polite number is obtained from the previous one by adding a constant.
 all the polite numbers are:





3, 5, 7, 9, 11, 13, 15, 17, …
6, 9, 12, 15, 18, …
10, 14, 18, …
15, 20, 25, …
21, 27, 33, …
Hence, the smallest number with a degree of politeness of 3 is 15.
19.
Find the value of √12 + √12 + √12+. . .
Solution
Let x = √12 + √12 + √12+. . .
Then x2 = 12 + √12 + √12+. . .
x2 = 12 + x
x2  x  12 = 0
(x + 3)(x  4) = 0
x = 4 or 3 (rejected because x is the positive square root)
 √12 + √12 + √12+. . . = 4
20.
The figure shows a circle with 4 points on its circumference. Each point is joined to
every other point by a line (called a chord). The chords divide the circle into 8 regions.
Find the maximum number of regions formed by the chords for a circle with 7 points.
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SASMO 2015, Secondary 4 Contest
Solution
The maximum number of regions occurs when no three chords intersect one another
inside the circle. Draw and count.
 maximum no. of regions formed by the chords for a circle with 7 points = 57.
𝑛
𝑛
𝑛−1
Note: The formula is not 2n1 = 271 = 64, but ( ) + (
) + ( ).
4
1
2
21.
Find the range of values of k for which the equation kx2  2015x + (k – 2015) = 0 has
one positive and one negative root.
Solution
Method 1
Let the two roots of the equation kx2  2015x + (k – 2015) = 0 be  and .
Since one of the roots is positive and the other negative, then  < 0.
Now  =
𝑐
𝑎
=
𝑘−2015
𝑘
< 0.
If k = 0, then the coefficient of x2 = 0, i.e. the equation will become a linear equation
with only one root, contradicting that there is a positive root and one negative root.
If k < 0, then k – 2015 > 0, i.e. k > 2015, which contradicts that k < 0 in the first place.
If k > 0, then k – 2015 < 0, i.e. k < 2015.  0 < k < 2015.
Method 2
The curve y = kx2  2015x + (k – 2015) cuts the x-axis at two values, one of which is
positive and the other negative.
If k = 0, then the coefficient of x2 = 0, i.e. the equation will become a linear equation
with only one root, contradicting that there is a positive root and one negative root.
If k < 0, then the coefficient of x2 will be negative, i.e. the curve will cut the y-axis at a
value that is positive, i.e. the y-intercept of the curve, k – 2015, will be positive.
This means that k > 2015, which contradicts that k < 0 in the first place.
If k > 0, then the coefficient of x2 will be positive, i.e. the curve will cut the y-axis at a
value that is negative, i.e. the y-intercept of the curve, k – 2015, will be negative.
This means that k < 2015.  0 < k < 2015.
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SASMO 2015, Secondary 4 Contest
22.
A circle and a triangle are drawn on a rectangular sheet of paper. What is the biggest
number of regions that can be formed on the paper?
Solution
To form the biggest number of regions, the triangle and the circle should intersect as
often as possible.
Since a line can cut a circle at most two times, then the following diagram shows the
biggest number of regions that can be formed between a triangle and a circle.
3
4
2
8
7
5
1
6
The next step is to divide the 8th region into as many regions as possible.
This is achieved by using the 3 vertices of the triangle, and the 3 curved parts of the
circle that are outside the triangle, to touch the edges of the rectangular sheet of
paper, as shown in the diagram below.
 biggest number of regions that can be formed on the paper = 13
3
10
9
4
2
7
11
1
12
23.
8
6
5
13
Find the sum of the coefficients in the expansion of 6 x 2  5 x  2  3  2 x  x 2  .
3
Solution
6 x

3

4
 5 x  2 3  2 x  x 2 = 𝑎0 + 𝑎1 𝑥 + 𝑎2 𝑥 2 +. . . + 𝑎14 𝑥14
Let x = 1. Then 𝑎0 + 𝑎1 + 𝑎2 +. . . + 𝑎14 = (6  5 + 2)3(3  2 + 1)4
= 33  24
= 432
2
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4
SASMO 2015, Secondary 4 Contest
24.
Albert and Bernard just become friends with Cheryl, and they want to know when her
birthday is. Cheryl gives them a list of 10 possible dates.
May 15
June 17
July 14
August 14
May 16
June 18
July 16
August 15
May 19
August 17
Cheryl then tells Albert and Bernard separately the month and the day of her birthday
respectively.
Albert:
I don’t know when Cheryl’s birthday is, but I know that Bernard does not
know too.
Bernard: At first I don’t know when Cheryl’s birthday is, but I know now.
Albert: Then I also know when Cheryl’s birthday is.
So when is Cheryl’s birthday?
Solution
Out of the 10 dates, the day ranges from 14 to 19, with only 18 and 19 each occurring
once. If the day of Cheryl’s birthday is 18 or 19, then Bernard would have known
when Cheryl’s birthday is since Cheryl has told him the day of her birthday.
But why does Albert know that Bernard does not know?
If Cheryl has told Albert that her birth month is May or June, then it is possible that
her birthday may be May 19 or June 18. This means that Bernard may know when
Cheryl’s birthday is. The fact that Albert knows that Bernard does not know means
that Cheryl has told Albert that her birth month is either July or August.
Initially, Bernard does not know when Cheryl’s birthday is, but how did he know after
Albert has first spoken?
Out of the 5 remaining dates in July and August, the day ranges from 15 to 17, with
only 14 occurring twice.
If Cheryl has told Bernard the day of her birthday is 14, then Bernard would not have
known. The fact that Bernard knows means the day of her birthday is not 14. So
now we are left with 3 possible dates: July 16, August 15 and August 17.
After Bernard has spoken, Albert now knows when Cheryl’s birthday is. If Cheryl has
told Albert her birth month is August, then Albert would not have known because
there are two possible dates in August.
 Cheryl’s birthday is on July 16.
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SASMO 2015, Secondary 4 Contest
25.
Find the last six digits of 20152015.
Solution
Since the last 6 digits of a product ab depends only on the last 6 digits of a and of b,
then
2015 = 002 015
Last 6 digits of 002 015  2015 = 060 225
Last 6 digits of 060 225  2015 = 353 375
Last 6 digits of 353 375  2015 = 050 625
Last 6 digits of 
Last 6 digits of 009 375  2015 = 890 625
Last 6 digits of 890 625  2015 = 609 375
Last 6 digits of 609 375  2015 = 890 625
 the last 6 digits repeat with a period of 2, with the exception of the first 5 powers.
Since the index 2015 is odd, then the last 6 digits of 152015 are 609 375.
40