Division
Singapore and Asian
P5
Schools Math Olympiad
2015
Full Name:
Index Number:
Class:
School:
SASMO 2015 Primary 5 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.
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SASMO 2015, Primary 5 Contest
SASMO 2015 Primary 5 [10 MCQ + 10 non-MCQ = 20 Q]
Starting Score = 10 marks (to avoid negative marks); Max Possible Score = 70 marks
Section A (Correct answer = 2 marks; no answer = 0; incorrect answer = minus 1 mark)
1.
An operator  acts on two numbers to give the following outcomes:
3  2 = 61
5  3 = 152
6  1 = 65
9  4 = 365
What is 7  5 equal to?
(a)
230
(b)
235
(c)
302
(d)
352
(e)
None of the above
_____________________________________________________________
2.
Fill in the blank: 2 tens 5 ones more than ____________ is 4 tens 3 ones.
(a)
18
(b)
28
(c)
58
(d)
68
(e)
None of the above
13
SASMO 2015, Primary 5 Contest
3.
Find the missing term in the following sequence: 1, 8, 27, 64, _____, 216.
(a)
81
(b)
100
(c)
125
(d)
149
(e)
181
____________________________________________________________
4.
Find the smallest whole number between 1 and 100 that is divisible by 12 and by
18.
(a)
(b)
(c)
(d)
(e)
24
36
48
54
72
14
SASMO 2015, Primary 5 Contest
5.
If the five-digit number 123N5 is divisible by 9, find N.
(a)
2
(b)
4
(c)
5
(d)
7
(e)
9
_____________________________________________________________________
6.
The diagram shows how an equilateral triangle can be cut into four pieces and
rearranged to form a square. This solution of the Haberdasher’s Puzzle is discovered by
Henry Dudeney (1857 – 1930).
If the length of the square is 13 cm and the height of the triangle is 17 cm, find
the length of the triangle, correct to the nearest whole number.
(a)
18 cm
(b)
19 cm
(c)
20 cm
(d)
21 cm
(e)
22 cm
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SASMO 2015, Primary 5 Contest
7.
24 means 2 multiplied by itself 4 times, i.e. 24 =  = 16. Find the last
digit of 510.
(a)
0
(b)
1
(c)
5
(d)
6
(e)
None of the above
_________________________________________________________________
8.
In a soccer tournament, 12 teams play against each other twice. How many
games are there altogether?
(a)
24
(b)
66
(c)
78
(d)
132
(e)
156
16
SASMO 2015, Primary 5 Contest
9.
A whole number multiplied by itself will give a special type of numbers called
perfect squares. Examples of perfect squares are 9 (= 3  3) and 16 (= 4  4). A
perfect square year is a year which is a perfect square. When is the next perfect
square year?
(a)
2016
(b)
2020
(c)
2025
(d)
2116
(e)
None of the above
_________________________________________________________________
10.
There are 4 items (a ruler, a pencil, an eraser and a sharpener) lying in a
straight row on a table. The sharpener is neither next to the ruler nor next to the
eraser. The pencil is not the third item. The eraser is neither the first nor the last
item. What is the order of the items on the table from first to last?
(a)
Pencil, eraser, ruler, sharpener
(b)
Sharpener, eraser, ruler, pencil
(c)
Pencil, eraser, sharpener, ruler
(d)
Pencil, ruler, eraser, sharpener
(e)
Sharpener, pencil, eraser and ruler
17
SASMO 2015, Primary 5 Contest
11.
The diagram shows a big rectangle that is made up of 4 small identical
rectangles. If the bigger dimension of one of the small rectangles is 20 cm, find
the area of the big rectangle.
(a)
600 cm2
(b)
800 cm2
(c)
2015 cm2
(d)
3200 cm2
(e)
Not enough given information to find
________________________________________________________________
12.
Which of the following statement(s) is or are correct?
Statement A: 8 + (0  2015) = 8
Statement B: 8 + (2015  0) = 8
Statement C: 8 + (0  2015) = 8
(a)
Only Statement C is correct.
(b)
Only Statements B and C are correct.
(c)
Only Statements A and C are correct.
(d)
All the three statements are correct.
(e)
None of the above
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SASMO 2015, Primary 5 Contest
13.
Two numbers x and y are such that

x is greater than or equal to 3, but less than or equal to 6.

y is greater than or equal to 8, but less than or equal to 10
Find the least possible value of y  x.
(a)
0
(b)
2
(c)
3
(d)
5
(e)
None of the above
_________________________________________________________________
14.
A palindromic number is a whole number that reads the same forward and
backward. For example, 1221 is a palindromic number. How many 3-digit
palindromic numbers are there?
(a)
19
(b)
90
(c)
100
(d)
900
(e)
None of the above
19
SASMO 2015, Primary 5 Contest
15.
Simplify the following fraction:
(a)
4
(b)
4
2+
3
4
9
11
(c) 1
1
(d) 2
3
(e)
1
4
4
None of the above
___________________________________________________________________
Section B (Correct answer = 4 marks; incorrect or no answer = 0)
16.
Cheryl wants to cut rectangular cards of length 4 cm by 3 cm from a rectangular
sheet 30 cm by 22 cm. Find the biggest number of cards that can be cut from
the sheet.
20
SASMO 2015, Primary 5 Contest
17.
The product of two numbers is 100. Neither of the two numbers has 10 as a
factor. Find the sum of these two numbers.
_________________________________________________________________
18.
A circle and a triangle are drawn on a flat surface. What is the biggest number of
regions that can be formed on the surface?
21
SASMO 2015, Primary 5 Contest
19.
A train travels at a speed of 60 km/h. The length of the train is 100 m. Find the
time taken by the train to pass completely through a 1-km tunnel.
_________________________________________________________________
20.
How many rectangles are there in a 3  3 square grid?
Note: A square is also a rectangle.
22
SASMO 2015, Primary 5 Contest
21.
What is the length of the largest cube that can be made from 2015 onecentimetre cubes?
________________________________________________________________
22.
Stationery Shop A has 50 pens while Stationery Shop B has 26 pens. After both
shops sold an equal number of pens, the ratio of pens that Shop A has to Shop B
is 3 : 1. Find the number of pens sold by each shop.
23
SASMO 2015, Primary 5 Contest
23.
The diagram shows a rectangle being divided into 3 smaller rectangles and a
square. If the perimeter of the unshaded rectangle is 36 cm and the area of the
square is 25 cm2, find the total area of the shaded rectangles.
_______________________________________________________________
24.
Given that 4! means  = 24, find the number of consecutive zeroes
at the end of 16!
24
SASMO 2015, Primary 5 Contest
25.
Find the remainder when 22015 is divided by 3.
End of Paper
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SASMO 2015, Primary 5 Contest
SOLUTION FOR SASMO 2015 PRIMARY 5
Section A
1.
An operator  acts on two numbers to give the following outcomes:
3
5
6
9




2
3
1
4
=
=
=
=
61
152
65
365
What is 7  5 equal to?
(a)
(b)
(c)
(d)
(e)
230
235
302
352 [Ans]
None of the above
Solution
a  b = (𝑎 × 𝑏)(𝑎 − 𝑏)
 7  5 = 352
2.
Fill in the blank: 2 tens 5 ones more than ____________ is 4 tens 3 ones.
(a)
(b)
(c)
(d)
(e)
18 [Ans]
28
58
68
None of the above
Solution
25 + ____ = 43
 the missing number is 43  25 = 18
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SASMO 2015, Primary 5 Contest
3.
Find the missing term in the following sequence: 1, 8, 27, 64, _____, 216.
(a)
(b)
(c)
(d)
(e)
81
100
125 [Ans]
149
181
Solution
Observe that 1 = 13, 8 = 23, 27 = 33, 64 = 43 and 216 = 63.
 the missing term is 53 = 125.
4.
Find the smallest whole number between 1 and 100 that is divisible by 12 and by
18.
(a)
(b)
(c)
(d)
(e)
24
36 [Ans]
48
54
72
Solution
Method 1
Numbers between 1 and 100 that are divisible by 18 are: 18, 36, 54, 72 and 90.
Of these 5 numbers, only 36 and 72 are divisible by 12.
 smallest whole number between 1 and 100 that is divisible by 12 and by 18 is
36.
Note: If we start with numbers divisible by 12, there will be more possibilities.
Method 2
A number that is exactly divisible by both 12 and 18 must also be exactly
divisible by 36.
The only numbers between 1 and 100 that are exactly divisible by 36 are 36 and
72.
 smallest whole number between 1 and 100 that is divisible by 12 and by 18 is
36.
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SASMO 2015, Primary 5 Contest
5.
If the five-digit number 123N5 is divisible by 9, find N.
(a)
(b)
(c)
(d)
(e)
2
4
5
7 [Ans]
9
Solution
Using the divisibility test for 9, 1 + 2 + 3 + N + 5 = N + 11 is also divisible by 9.
Since N is a single digit, N = 7.
6.
The diagram shows how an equilateral triangle can be cut into four pieces and
rearranged to form a square. This solution of the Haberdasher’s Puzzle is
discovered by Henry Dudeney (1857 – 1930).
If the length of the square is 13 cm and the height of the triangle is 17 cm, find
the length of the triangle, correct to the nearest whole number.
(a)
(b)
(c)
(d)
(e)
18 cm
19 cm
20 cm [Ans]
21 cm
22 cm
Solution
Area of square = 13  13 = 169 cm2
1
1
Area of triangle = 2  base  height = 2  length of triangle  17 cm
Since both areas are equal, then length of triangle
= 169  2  17
= 20 cm (to nearest whole number)
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SASMO 2015, Primary 5 Contest
7.
24 means 2 multiplied by itself 4 times, i.e. 24 = 2  2  2  2 = 16. Find the last
digit of 510.
(a)
(b)
(c)
(d)
(e)
0
1
5 [Ans]
6
None of the above
Solution
5=5
5  5 = 25
25  5 = 125
Since the last digit of a product ab depends only on the last digit of a and of b,
then the last digit of 510 is 5.
8.
In a soccer tournament, 12 teams play against each other twice. How many
games are there altogether?
(a)
(b)
(c)
(d)
(e)
24
66
78
132 [Ans]
156
Solution
Consider the 12 teams playing against each other once first.
The first team will play against 11 other teams, i.e. 11 games;
the second team will play against 10 other teams, i.e. 10 games;
the third team will play against 9 other teams, i.e. 9 games; etc.
Thus total no. of games = 11 + 10 + 9 + … + 3 + 2 + 1
1 + 11 = 12
2 + 10 = 12
3 + 9 = 12
5 pairs
4 + 8 = 12
5 + 7 = 12
6
 total no. of games = 12  5 + 6 = 66
Since the 12 teams play against each other twice, total no. of games
= 66  2 = 132
29
SASMO 2015, Primary 5 Contest
9.
A whole number multiplied by itself will give a special type of numbers called
perfect squares. Examples of perfect squares are 9 (= 3  3) and 16 (= 4  4). A
perfect square year is a year which is a perfect square. When is the next perfect
square year?
(a)
(b)
(c)
(d)
(e)
2016
2020
2025 [Ans]
2116
None of the above
Solution
Since 40  40 = 1600 and 50  50 = 2500, we will have to try the square of a
whole number between 40 and 50.
Since 44  44 = 1936 and 45  45 = 2025, then the next perfect square year is
2025.
30
SASMO 2015, Primary 5 Contest
10.
There are 4 items (a ruler, a pencil, an eraser and a sharpener) lying in a
straight row on a table. The sharpener is neither next to the ruler nor next to the
eraser. The pencil is not the third item. The eraser is neither the first nor the last
item. What is the order of the items on the table from first to last?
(a)
(b)
(c)
(d)
(e)
Pencil, eraser, ruler, sharpener
Sharpener, eraser, ruler, pencil
Pencil, eraser, sharpener, ruler
Pencil, ruler, eraser, sharpener
Sharpener, pencil, eraser and ruler [Ans]
Solution
Let R = Ruler, P = Pencil, E = Eraser and S = Sharpener.
The eraser is neither the first nor the last item.
Suppose the eraser is the second item:
_____, __E__, _____, _____
The sharpener is neither next to the ruler nor next to the eraser.
This means that the sharpener is the last item, and the ruler is the first item:
__R__, __E__, _____, __S__
So the pencil is the third item, which contradicts the given statement that the
pencil is not the third item.
Thus the eraser must be the third item:
_____, _____, __E__, _____
The sharpener is neither next to the ruler nor next to the eraser.
This means that the sharpener is the first item, and the ruler is the last item:
__S__, _____, __E__, __R__
So the pencil ball is the second item.
 the order of the items on the table from first to last is sharpener, pencil,
eraser and ruler.
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SASMO 2015, Primary 5 Contest
11.
The diagram shows a big rectangle that is made up of 4 small identical
rectangles. If the bigger dimension of one of the small rectangles is 20 cm, find
the area of the big rectangle.
(a)
(b)
(c)
(d)
(e)
600 cm2
800 cm2 [Ans]
2015 cm2
3200 cm2
Not enough given information to find
Solution
Since the bigger dimension of the small rectangle is 20 cm, then the smaller
dimension of the small rectangle is 10 cm as shown:
20
10
10
So the length of the big rectangle is 10 + 20 + 10 = 40 cm
 area of big rectangle = 40 × 20 = 800 cm2
12.
Which of the following statement(s) is or are correct?
Statement A: 8 + (0  2015) = 8
Statement B: 8 + (2015  0) = 8
Statement C: 8 + (0 × 2015) = 8
(a)
(b)
(c)
(d)
(e)
Only Statement C is correct.
Only Statements B and C are correct.
Only Statements A and C are correct. [Ans]
All the three statements are correct.
None of the above
Solution
Statement A: 8 + (0  2015) = 8 + 0 = 8
Statement B: 8 + (2015  0) is undefined.
Statement C: 8 + (0 × 2015) = 8 + 0 = 8
 only Statements A and C are correct.
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SASMO 2015, Primary 5 Contest
13.
Two numbers x and y are such that


x is greater than or equal to 3, but less than or equal to 6.
y is greater than or equal to 8, but less than or equal to 10
Find the least possible value of y  x.
(a)
(b)
(c)
(d)
(e)
0
2 [Ans]
3
5
None of the above
Solution
Least possible value of y  x
= least possible value of y  greatest possible value of x
=86
=2
14.
A palindromic number is a whole number that reads the same forward and
backward. For example, 1221 is a palindromic number. How many 3-digit
palindromic numbers are there?
(a)
(b)
(c)
(d)
(e)
19
90 [Ans]
100
900
None of the above
Solution
Method 1
The hundreds digit of a 3-digit number cannot be 0.
If the hundreds digit of a 3-digit palindromic number is 1, there are 10 such
numbers: 101, 111, 121, …, 191.
Similarly, if the hundreds digit of a 3-digit palindromic number is 2, there are
also 10 such numbers: 202, 212, 222, …, 292.
Since the hundreds digit of a 3-digit palindromic number can be 1, 2, 3, … or 9,
then there are 9 × 10 = 90 three-digit palindromic numbers.
33
SASMO 2015, Primary 5 Contest
Method 2
There are only 9 possibilities for the hundreds digit because the hundreds digit of
a 3-digit number cannot be 0.
Since the ones digit is the same as the hundreds digit in a 3-digit palindromic
number, there are no other possibilities for the ones digit once the hundreds
digit is fixed.
However, there are 10 possibilities for the tens digit.
 there are 9 × 10 = 90 three-digit palindromic numbers.
15.
Simplify the following fraction:
(a)
(b)
4
𝟒
.
3
4
[Ans]
𝟏𝟏
1
(d) 2
3
(e)
2+
9
1
(c)
1
4
4
None of the above
Solution
1
3
2+
4
=
1
8 3
+
4 4
=
1
11
4
=
𝟒
𝟏𝟏
34
SASMO 2015, Primary 5 Contest
Section B
16.
Cheryl wants to cut rectangular cards of length 4 cm by 3 cm from a rectangular
sheet 30 cm by 22 cm. Find the biggest number of cards that can be cut from
the sheet.
Solution
There are 3 ways to cut.
First way to cut:
1
30 cm  4 cm = 72, i.e. 7 cards along the length of the rectangular sheet.
1
22 cm  3 cm = 73, i.e. 7 cards along the breadth of the rectangular sheet.
Total number of cards from the first way to cut = 7 × 7 = 49.
Second way to cut:
30 cm  3 cm = 10 cards along the length of the rectangular sheet.
1
22 cm  4 cm = 5 , i.e. 5 cards along the breadth of the rectangular sheet.
2
Total number of cards from the second way to cut = 10 × 5 = 50.
Third way to cut:
Cut the rectangular sheet 30 cm by 22 cm into two smaller rectangular sheets 30
cm by 16 cm and 30 cm by 6 cm.
The first rectangular sheet 30 cm by 16 cm can be cut into
40 cards.
30
3
×
16
4
= 10 × 4 =
Cut the second rectangular sheet 30 cm by 6 cm into two even smaller
rectangular sheets 28 cm by 6 cm and 2 cm by 6 cm.
The rectangular sheet 28 cm by 6 cm can be cut into
28
4
6
× = 7 × 2 = 14 cards.
3
The rectangular sheet 2 cm by 6 cm cannot be cut into any more cards.
Total number of cards from the third way to cut = 40 + 14 = 54
 biggest number of cards that can be cut from the sheet = 54
35
SASMO 2015, Primary 5 Contest
17.
The product of two numbers is 100. Neither of the two numbers has 10 as a
factor. Find the sum of these two numbers.
Solution
100 = 2 × 2 × 5 × 5
Since 2 × 5 = 10, each of the two numbers must not contain both a 2 and a 5 as
its factors, or else it will contain 10 as its factor.
Thus 100 = (2 × 2) × (5 ×5) = 4 × 25.
 the two numbers are 4 and 25, and their sum is 4 + 25 = 29.
18.
A circle and a triangle are drawn on a flat surface. What is the biggest number of
regions that can be formed on the surface?
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
that the biggest number of regions that can be formed on the surface is 8.
3
4
2
8
7
5
1
6
19.
A train travels at a speed of 60 km/h. The length of the train is 100 m. Find the
time taken by the train to pass completely through a 1-km tunnel.
Solution
Total distance travelled by the train when it passes completely through the
tunnel
= 1 km + 100 m = 1.1 km
Speed of train = 60 km/h = 1 km/min
Total time taken =
Distance
Speed
=
1.1
1
= 1.1 min or 1 min 6 s
36
SASMO 2015, Primary 5 Contest
20.
How many rectangles are there in a 3  3 square grid?
Note: A square is also a rectangle.
Solution
Method 1
A square is also a rectangle.
Type of Rectangles
1 × 1 square
2 × 2 square
3 × 3 square
1 × 2 rectangle
2 × 1 rectangle
1 × 3 rectangle
3 × 1 rectangle
2 × 3 rectangle
3 × 2 rectangle
Total No.
Number of Rectangles
9
4
1
6
6
3
3
2
2
36
Method 2
There is exactly one rectangle for every pair of distinct horizontal lines and every
pair of distinct vertical lines.
 total no. of rectangles in a 3  3 square grid
= no. of pairs of distinct horizontal lines  no. of pairs of distinct vertical lines
=
4×3
2
=66
= 36
21.

4×3
2
What is the length of the largest cube that can be made from 2015 onecentimetre cubes?
Solution
Since 12 × 12 × 12 = 1728 and 13 × 13 × 13 = 2197, then the length of the
largest cube that can be made from 2015 one-centimetre cubes is 12 cm.
37
SASMO 2015, Primary 5 Contest
22.
Stationery Shop A has 50 pens while Stationery Shop B has 26 pens. After both
shops sold an equal number of pens, the ratio of pens that Shop A has to Shop B
is 3 : 1. Find the number of pens sold by each shop.
Solution
Method 1 (Model Method)
After
Shop A
Shop B
50
Before
Shop A
Shop B
26
From the Before model, 2 units = 50  26 = 24
1 unit = 12
 from Shop B, no. of pens sold by each shop = 26  12 = 14
Note: Alternatively, triple the boxes in Shop B in the Before model, and then
subtract the boxes in Shop A to find 2 shaded units = 26 × 3  50 = 28.
Method 2 (Algebra)
Let the no. of pens sold by each shop be x.
Then 50  x = 3(26  x)
50  x = 78  3x
2x = 28
x = 14
 no. of pens sold by each shop = 14
38
SASMO 2015, Primary 5 Contest
23.
The diagram shows a rectangle being divided into 3 smaller rectangles and a
square. If the perimeter of the unshaded rectangle is 36 cm and the area of the
square is 25 cm2, find the total area of the shaded rectangles.
Solution
Put the two shaded rectangles to form a long rectangle as shown:
Length of long rectangle =
=
1
2
1
2
× perimeter of unshaded rectangle
× 36 cm
= 18 cm
Breadth of long rectangle = length of square = 5 cm
 total area of shaded rectangles = 18 cm × 5 cm = 90 cm2
24.
Given that 4! means 4 × 3 × 2 × 1 = 24, find the number of consecutive zeros
at the end of 16!
Solution
Notice that 2  5 = 10, 22  52 = 100, etc., i.e. a zero at the end is produced by
the product of a factor of 2 and a factor of 5.
16!, when written as a product of prime factors, will contain three 5s and more
than three 2s among its factors.
 there are at most 3 pairs of factor 2 and factor 5, i.e. there are 3 consecutive
zeros at the end of 16!
39
SASMO 2015, Primary 5 Contest
25.
Find the remainder when 22015 is divided by 3.
Solution
Observe the following pattern: when divided by 3,
21 leaves a remainder of 2,
22 = 4 leaves a remainder of 1,
23 = 8 leaves a remainder of 2,
24 = 16 leaves a remainder of 1, …
This means that the remainder will repeat with a period of 2.
Since 2015 is odd, then 22015, when divided by 3, will leave a remainder of 2.
Note: You need to ensure there is a good reason why the above pattern will
continue:
22 = 4 = 3 + 1 = 1 group of 3 + 1
23 = 2 × 22 = 2 × (3 + 1) = 2 groups of 3 + 2
24 = 2 × 23 = 2 × (2 groups of 3 + 2) = 4 groups of 3 + 4
= 4 groups of 3 + 3 + 1
= 5 groups of 3 + 1
In other words, whenever you multiply by 2, all the groups of 3 will still be
divisible by 3.
If the remainder is 1, the next remainder will just be 2 × 1 = 2; if the remainder
is 2, the next remainder will be 1 since 2 × 2 = 4 = 3 + 1.
40