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SASMO 2014 Round 1 Primary 2 Problems
Section A [1 mark for each question]
1.
What is 2014 + 2  0  1  4 equal to?
(a) 2014
(b) 2016
(c) 2021
(d) 2022
(e) None of the above
2.
Ten lampposts are equally spaced along a straight line. The distance between two
consecutive lampposts is 40 m. What is the distance between the first and the last
lampposts?
(a) 360 m
(b) 380 m
(c) 400 m
(d) 420 m
(e) None of the above
3.
Find the next term of the following sequence: 1, 1, 2, 3, 5, …
(a) 6
(b) 7
(c) 8
(d) 9
(e) 10
4.
Jane wrote the word STUDENTS thrice. How many times did she write the letter S?
(a) 2
(b) 4
(c) 6
(d) 8
(e) 10
5.
What number between 37 and 47 is exactly divisible by both 2 and 3?
(a) 38
(b) 39
(c) 42
(d) 44
(e) 45
6.
A shop sells sweets where every 3 sweet wrappers can be exchanged for one more
sweet. Ali has enough money to buy only 7 sweets. What is the biggest number of
sweets that he can get from the shop?
(a) 7
(b) 8
(c) 9
(d) 10
(e) 11
7.
There are 14 children playing “The eagle catches the chicks.” One of them is the
‘eagle’ while another child is the ‘mother hen’ whose job is to protect the ‘chicks’.
The rest of the children are the ‘chicks’. After a while, the ‘eagle’ has caught 5
‘chicks’. How many ‘chicks’ are still running around?
(a) 6
(b) 7 (c) 8 (d) 9 (e) 10
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SASMO 2014 Round 1 Primary 2 Solutions
Section A [1 mark for each question]
1.
What is 2014 + 2  0  1  4 equal to?
(a) 2014
(b) 2016
(c) 2021
(d) 2022
(e) None of the above
Solution
2014 + 2  0  1  4 = 2014 + 0 = 2014 (a)
2.
Ten lampposts are equally spaced along a straight line. The distance between two
consecutive lampposts is 40 m. What is the distance between the first and the last
lampposts?
(a) 360 m
(b) 380 m
(c) 400 m
(d) 420 m
(e) None of the above
Solution
Distance between the first and the last lampposts = 40 m  9 gaps = 360 m (a)
3.
Find the next term of the following sequence: 1, 1, 2, 3, 5, …
(a) 6
(b) 7
(c) 8
(d) 9
(e) 10
Solution
From the third term onwards, the next term is obtained by adding the previous two
terms.
 the next term is 3 + 5 = 8 (c)
4.
Jane wrote the word STUDENTS thrice. How many times did she write the letter S?
(a) 2
(b) 4
(c) 6
(d) 8
(e) 10
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Solution
‘Thrice’ means 3 times.
 Jane wrote the letter S a total of 2  3 = 6 times (c)
5.
What number between 37 and 47 is exactly divisible by both 2 and 3?
(a) 38
(b) 39
(c) 42
(d) 44
(e) 45
Solution
Method 1
Numbers between 37 and 47 that are exactly divisible by 2 are: 38, 40, 42, 44 and 46.
Of these 5 numbers, only 42 is exactly divisible by 3.
 the number between 37 and 47 that is exactly divisible by both 2 and 3 is 42 (c).
Method 2
A number that is exactly divisible by both 2 and 3 must also be exactly divisible by 6.
he only number between 37 and 47 that is exactly divisible by 6 is 42.
 the number between 37 and 47 that is exactly divisible by both 2 and 3 is 42 (c).
6.
A shop sells sweets where every 3 sweet wrappers can be exchanged for one more
sweet. Ali has enough money to buy only 7 sweets. What is the biggest number of
sweets that he can get from the shop?
(a) 7
(b) 8
(c) 9
(d) 10
(e) 11
Solution
7 sweets  7 wrappers  2 sweets and 1 wrapper  3 wrappers  1 sweet
 biggest no. of sweets = 7 + 2 + 1 = 10 (d)
7.
There are 14 children playing “The eagle catches the chicks.” One of them is the
‘eagle’ while another child is the ‘mother hen’ whose job is to protect the ‘chicks’.
The rest of the children are the ‘chicks’. After a while, the ‘eagle’ has caught 5
‘chicks’. How many ‘chicks’ are still running around?
(a) 6
(b) 7
(c) 8
(d) 9
(e) 10
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Solution
No. of ‘chicks’ still running around = 14  1 (eagle)  1 (mother hen)  5 = 7 (b)
8.
Find the number A such that the following statement is true: 7  A = 3  8 + 4  8.
(a) 3
(b) 4
(c) 5
(d) 7
(e) 8 [Ans]
Solution
Method 1
7  A = 3  8 + 4  8 = 24 + 32 = 56
 A = 56  7 = 8 (e)
Method 2
7  A = 3  8 + 4  8 = (3 + 4)  8 = 7  8
 A = 8 (e)
9.
Two $1 coins and ten 50¢ coins are randomly distributed among 4 children such that
each child receives the same number of coins. What is the difference between the
biggest amount and the smallest amount a child can receive?
(a) 50¢
(b) $1
(c) $1.50
(d) $2
(e) None of the above
Solution
There are a total of 2 + 10 = 12 coins.
So each child receives 12  4 = 3 coins.
Biggest amount a child can receive = $1 + $1 + 50¢ = $2.50
Smallest amount a child can receive = 50¢ + 50¢ + 50¢ = $1.50
 difference between biggest amount and smallest amount = $2.50  $1.50 = $1 (b)
10.
Tim is 8 years old and Sally is 4 years old. How old will Sally be when Tim is 14
years old?
(a) 7
(b) 8
(c) 9
(d) 10
(e) None of the above
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Solution
Method 1
Tim will be 14 years old in 14  8 = 6 years’ time.
 Sally will be 6 + 4 = 10 years old (d).
Method 2
Difference in age between Tim and Sally = 8  4 = 4 years
 when Tim is 14 years old, Sally will be 14  4 = 10 years old (d).
Section B [3 marks for each question]
11.
In the following alphametic, all the different letters stand for different digits. Find P
and I.
I
I
+
I
P
I
Solution
I  3 = _I
By guess and check, the only possible solution for I is 5.
 5 + 5 + 5 = 15, i.e. P = 1 and I = 5.
12.
A box contains 4 balls of different colours (red, green, yellow and blue) lying in a
row. The green ball is not the second ball. The red ball is neither the first nor the last
ball. The yellow ball is neither next to the red ball nor next to the blue ball. What is
the order of the balls in the box from first to last?
Solution
The red ball is neither the first nor the last ball.
Suppose the red ball is the third ball:
_____, _____, __R__ , _____
The yellow ball is neither next to the red ball nor next to the blue ball.
This means that the yellow ball is the first ball, and the blue ball is the last ball:
__Y__, _____, __R__ , __B__
So the green ball is the second ball, which is a contradiction.
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Method 3
The pattern for the no. of white squares is 3  3  1, 3  4  2, 3  5  3, … (total no.
of squares minus no. of black squares in each figure)
 no. of white squares that will surround one row of 50 black squares = 3  52  50
= 106 (e)
Method 4
The pattern for the no. of white squares is 8, 10, 12, …, which is equal to 2  4, 2  5,
2  6, …
 no. of white squares that will surround one row of 50 black squares = 2  53
= 106 (e)
15.
The diagram shows 9 points. Draw 4 consecutive line segments (i.e. the start point of
the next segment must coincide with the endpoint of the previous segment) to pass
through all the 9 points.
Solution
If you try to draw the line segments within the region bounded by the dots, you will
realise that you need at least 5 consecutive line segments.
 you must draw some of the line segments outside the region as shown:
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SASMO 2014 Round 1 Primary 3 Problems
Section A [1 mark for each question]
1.
Jane is 9 years old and John is 5 years old. How old will John be when Jane is 15
years old?
2.
A textbook is opened at random. To what pages is it opened if the product of the
facing pages is 110?
3.
Find the number B such that the following statement is true: 8  B = 3  9 + 5  9.
4.
It is given that a  b = a  b + a  b. For example, 2  3 = 2  3 + 2  3 = 5. Find the
value of 4  3  3  4.
5.
Jane has a rope of length 23 cm. She wants to cut the rope so that she can form the
biggest possible square, where the length of each side, in cm, is a whole number.
What is the length of the rope that she must cut to form the square?
6.
Find the missing term in the following sequence: 1, 2, 6, 24, _____, 720.
7.
On National Day, 39 soldiers lined up in a straight row on opposite sides of Stadium
Street to welcome Prime Minister Lee. A soldier stands on each end of Stadium
Street. The distance between two adjacent soldiers on either side was 20 m. The
soldiers on one side were arranged such that each soldier filled the gap between two
other soldiers on the opposite side. How long was Stadium Street?
8.
A shop sells sweets where every 3 sweet wrappers can be exchanged for one more
sweet. Sharon has enough money to buy only 11 sweets. What is the biggest number
of sweets that she can get from the shop?
9.
At a workshop, there are 10 participants. Each of them shakes hand once with one
another. How many handshakes are there?
10.
Ali uses identical square tiles to make the following figures. If he continues using the
same pattern, how many tiles will there be in the 15th figure?
11.
What is the least number of cuts required to cut 16 identical sausages so that they can
be shared equally among 24 people?
12.
A vending machine accepts 10¢ coins, 20¢ coins, 50¢ coins and $1 coins only. Ivy
wants to buy a can of drinks that costs $1.60. She has eight 10¢ coins, three 20¢ coins,
two 50¢ coins and one $1 coin. If she wants to get rid of as many coins as possible,
what is the combination of coins that she should put inside the vending machine?
13.
The total cost of a pen and a pencil is $2.90. The pen costs 60¢ more than the pencil.
What much does the pen cost?
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14.
If the three-digit number 3N3 is divided by 9, the remainder is 1. Find N.
15.
Charles has 16 marbles. He divides them into 4 piles so that each pile has a different
number of marbles. Find the smallest possible number of marbles in the biggest pile.
Section B [2 marks for each question]
16.
In the following alphametic, all the different letters stand for different digits. Find the
three-digit sum SEE.
A
+
S
A
S
E
E
17.
Find the total number of triangles in the diagram.
18.
A teacher has a bag of sweets to treat her class. If she gave 5 sweets to each student,
then she would have 40 sweets left. If she gave 7 sweets to each student, then she
would have 6 sweets left. How many students and how many sweets are there?
19.
What are the last 2 digits of the sum 1 + 11 + 111 + … + 111…111?
50 digits
20.
Alvin tells the truth on Monday, Tuesday, Wednesday and Thursday. He lies on all
other days. Doris tells the truth on Monday, Friday, Saturday and Sunday. She lies on
all other days. One day they both said, “Yesterday I lied.” When was that ‘one day’?
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SASMO 2014 Round 1 Primary 3 Solutions
Section A [1 mark for each question]
1.
Jane is 9 years old and John is 5 years old. How old will John be when Jane is 15
years old?
Solution
Method 1
Jane will be 15 years old in 15  9 = 6 years’ time.
 John will be 6 + 5 = 11 years old.
Method 2
Difference in age between Jane and John = 9  5 = 4 years
 when Jane is 15 years old, John will be 15  4 = 11 years old.
2.
A textbook is opened at random. To what pages is it opened if the product of the
facing pages is 110?
Solution
Since 10  10 = 100, try 10  11 = 110.
 the pages are 10 and 11.
3.
Find the number B such that the following statement is true: 8  B = 3  9 + 5  9.
Solution
Method 1
8  B = 3  9 + 5  9 = 27 + 45 = 72
 A = 72  8 = 9
Method 2
8  B = 3  9 + 5  9 = (3 + 5)  9 = 8  9
A=9
4.
It is given that a  b = a  b + a  b. For example, 2  3 = 2  3 + 2  3 = 5. Find the
value of 4  3  3  4.
Solution
Method 1
4  3 = 4  3 + 4  3 = 13
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3  4 = 3  4 + 3  4 = 11
 4  3  3  4 = 13  11 = 2
Method 2
43=43+43=43+1
34=34+34=341
 4  3  3  4 = 1  (1) = 2
5.
Jane has a rope of length 23 cm. She wants to cut the rope so that she can form the
biggest possible square, where the length of each side, in cm, is a whole number.
What is the length of the rope that she must cut to form the square?
Solution
Since 5 cm  4 = 20 cm, and 6 cm  4 = 24 cm, then the biggest possible square that
she can form has a length of 5 cm.
 length of rope that she must cut to form the square = 5 cm  4 = 20 cm
6.
Find the missing term in the following sequence: 1, 2, 6, 24, _____, 720.
Solution
The pattern is as follows:
1,
2
2,
6,
3
24, _____, 720
4
5
6
 the missing term is 24  5 = 120.
7.
On National Day, 39 soldiers lined up in a straight row on opposite sides of Stadium
Street to welcome Prime Minister Lee. A soldier stands on each end of Stadium
Street. The distance between two adjacent soldiers on either side was 20 m. The
soldiers on one side were arranged such that each soldier filled the gap between two
other soldiers on the opposite side. How long was Stadium Street?
Solution
There are 20 soldiers on one side and 19 soldiers on the other side.
 length of Stadium Street = 20 m  (20 – 1) gaps = 380 m
8.
A shop sells sweets where every 3 sweet wrappers can be exchanged for one more
sweet. Sharon has enough money to buy only 11 sweets. What is the biggest number
of sweets that she can get from the shop?
Solution
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11 sweets  11 wrappers  3 sweets and 2 wrappers  5 wrappers  1 sweet and 2
wrappers  3 wrappers  1 sweet
 biggest no. of sweets = 11 + 3 + 1 + 1 = 16 [Common mistakes: 14, 15]
9.
At a workshop, there are 10 participants. Each of them shakes hand once with one
another. How many handshakes are there?
The first participant will shake hand with 9 other participants;
the second participant will shake hand with 8 other participants;
the third participant will shake hand with 7 other participants; etc.
Thus total no. of handshakes = 9 + 8 + 7 + … + 3 + 2 + 1
1 + 9 = 10
2 + 8 = 10
3 + 7 = 10
4 + 6 = 10
5
4 pairs
 total no. of handshakes = 10  4 + 5 = 45
10.
Ali uses identical square tiles to make the following figures. If he continues using the
same pattern, how many tiles will there be in the 15th figure?
Solution
Method 1
The two corner tiles are the same for all figures.
 the 15th figure will have 15  3 + 2 = 47 tiles.
Method 2
The tiles in the top row have this pattern: 3, 4, 5, 6, …
 the 15th figure will have 17 + 2  15 = 47 tiles.
Method 3
The tiles in each of the vertical column have this pattern: 2, 3, 4, 5, …
 the 15th figure will have 16  2 + 15 = 47 tiles.
Method 4
The no. of tiles in each figure is equal to the “area of the rectangle” minus the “area of
the hole in the middle”.
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The pattern for the “area of the rectangle” is 2  3, 3  4, 4  5, 5  6, …
The pattern for the “area of the hole in the middle” is 1  1, 2  2, 3  3, 4  4, …
 the 15th figure will have 16  17  15  15 = 272  225 = 47 tiles.
11.
What is the least number of cuts required to cut 16 identical sausages so that they can
be shared equally among 24 people?
Solution
Fraction of sausage each person will get =
=
This means that there must be at least 16 cuts since no one will get one whole
sausage.
Cut each of the 16 sausages at the -mark. Then 16 people will get one
sausage
each, and the remaining 8 people will get two sausages each.
 least no. of cuts = 16
12.
A vending machine accepts 10¢ coins, 20¢ coins, 50¢ coins and $1 coins only. Ivy
wants to buy a can of drinks that costs $1.60. She has eight 10¢ coins, three 20¢ coins,
two 50¢ coins and one $1 coin. If she wants to get rid of as many coins as possible,
what is the combination of coins that she should put inside the vending machine?
Solution
To get rid of as many coins as possible, we try to use as many coins with the smallest
value (i.e. the 10¢ coins) as possible.
If Ivy uses all the eight 10¢ coins, then what is left is $1.60  80¢ = 80¢.
Unfortunately, Ivy has only three 20¢ coins, which is not enough.
So she has to use a 50¢ coin.
But 50¢ + 20¢ = 70¢, which is 10¢ short.
In other words, she can’t use all the eight 10¢ coins.
 she has to use seven 10¢ coins, one 50¢ coin and two 20¢ coins.
13.
The total cost of a pen and a pencil is $2.90. The pen costs 60¢ more than the pencil.
What much does the pen cost?
Solution
Method 1 (Systematic Guess and Check)
Make a systematic list, starting with a random guess: pencil costs $1 and pen costs
$1.60.
Cost of Pencil
$1
$1.10
$1.15
Cost of Pen
$1.60
$1.70
$1.75
Total Cost
$2.60
$2.80
$2.90
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 the pen costs $1.75
Method 2 (Model Method)
Pen
60¢
$2.90
Pencil
2 units = $2.90  60¢ = $2.30
1 unit = $1.15
 the pen costs $1.15 + 60¢ = $1.75
Method 3 (Algebraic Method)
Let the cost of the pencil be $x.
Then the cost of the pen is $(x + 0.6).
So x + (x + 0.6) = 2.9
2x + 0.6 = 2.9
2x = 2.3
x = 1.15
 the pen costs $(1.15 + 0.60) = $1.75
14.
If the three-digit number 3N3 is divided by 9, the remainder is 1. Find N.
Solution
Since 3N3 gives a remainder of 1 when divided by 9, then 3N3  1 = 3N2 is divisible
by 9.
Using the divisibility test for 9, 3 + N + 2 = N + 5 is also divisible by 9.
 N = 4.
15.
Charles has 16 marbles. He divides them into 4 piles so that each pile has a different
number of marbles. Find the smallest possible number of marbles in the biggest pile.
Solution
For each pile to have a different number of marbles, and the biggest pile to have the
smallest possible number of marbles, put 1 marble in the 1st pile, 2 marbles in the
2nd pile, 3 marbles in the 3rd pile and 4 marbles in the 4th pile. So the biggest pile is
the 4th pile, but there are only 1 + 2 + 3 + 4 = 10 marbles.
The 11th marble will have to go to the 4th pile so that each pile will have a different
number of marbles. The 12th marble cannot go to the 4th pile because we want to
find the smallest possible number of marbles in the biggest pile, so the 12th marble
will have to go to the 3rd pile. Similarly, the 13th and 14th marbles will go to the 2nd
and 1st piles respectively.
The 15th marble will go to the 4th pile again, and the 16th marble to the 3rd pile.
 the largest pile (which is the 4th pile) will contain 4 + 1 + 1 = 6 marbles.
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Section B [2 marks for each question]
16.
In the following alphametic, all the different letters stand for different digits. Find the
three-digit sum SEE.
A
+
S
A
S
E
E
Solution
Since the addition of two digits will give a maximum of 18, or a maximum of 19 if
there is a carryover (or renaming) of 1, this means that the maximum carryover is
1.
So A = 9 and there is a carryover of 1 for A to give 10, i.e. S = 1 and E = 0.
 91 + 9 = 100, i.e. the three-digit sum SEE is 100.
17.
Find the total number of triangles in the diagram.
1
Solution
Label the vertices as shown in the diagram.
These are the vertices that form a triangle:
123, 124, 125, 126, 134, 135, 145, 156, 157, 158,
167, 168, 236, 256, 356, 457, 458, 568, 578, 678
 there is a total of 20 triangles in the diagram.
2
3
4
5
7
8
6
18.
A teacher has a bag of sweets to treat her class. If she gave 5 sweets to each student,
then she would have 40 sweets left. If she gave 7 sweets to each student, then she
would have 6 sweets left. How many students and how many sweets are there?
Solution
If the teacher gave 5 sweets to each student, then she would have 40 sweets left.
From the 40 sweets left, if she gave 2 more sweets to each student so that each student
has 7 sweets, then she would have 6 left.
This means that she gave a total of 40  6 = 34 sweets from the 40 sweets left to the
students.
Since each student receives only 2 more sweets (i.e. the 6th and 7th sweets), then there
are 34  2 = 17 students.
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19.
 there are 17  7 + 6 = 125 sweets.
What are the last 2 digits of the sum 1 + 11 + 111 + … + 111…111?
50 digits
Solution
Last digit: 50  0 (and carry 5)
Second last digit: 49 + 5 (carry) = 54  4
 the last 2 digits are 40.
20.
Alvin tells the truth on Monday, Tuesday, Wednesday and Thursday. He lies on all
other days. Doris tells the truth on Monday, Friday, Saturday and Sunday. She lies on
all other days. One day they both said, “Yesterday I lied.” When was that ‘one day’?
Solution
Mon
Tue
Wed
Thur




Alvin

Doris
 indicates the person telling the truth
Fri
Sat
Sun



If Alvin tells the truth on that ‘one day’ that he lied ‘yesterday’, then that ‘one day’
must be Monday.
If Alvin tells the lie on that ‘one day’ that he lied ‘yesterday’, then he must be telling
the truth ‘yesterday’ and so that ‘one day’ must be Friday.
 based on Alvin, it has to be either Monday or Friday.
If Carol tells the truth on that ‘one day’ that she lied ‘yesterday’, then that ‘one day’
must be Friday.
If Carol tells the lie on that ‘one day’ that she lied ‘yesterday’, then she must be
telling the truth ‘yesterday’ and so that ‘one day’ must be Tuesday.
 based on Carol, it has to be either Tuesday or Friday.
Hence, that ‘one day’, when they both said that they lied ‘yesterday’, has to be Friday.
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13.
Find the total number of squares in a 4  4 square grid.
14.
How many digits are there before the fifteen 7 of the following number?
37337333733337333337…
15.
A box contains 30 coloured pens: 15 blue, 10 red and 5 black. Alice takes some pens
from the box without looking at the colours of the pens. What is the least number of
pens she must take so that she has at least 8 pens of the same colour?
Section B [2 marks for each question]
16.
In the following alphametic, all the different letters stand for different digits. Find the
four-digit sum PEEL.
+
P
A
M
L
A
P
E
E
L
17.
A teacher has a bag of sweets to treat her class. If she gave 4 sweets to each student,
then she would have 82 sweets left. If she gave 9 sweets to each student, then she
would have 2 sweets left. How many students and how many sweets are there?
18.
When Amy, Betty and Cheryl eat out, each orders either beef or chicken.
a. If Amy orders beef, Betty orders chicken.
b. Either Amy or Cheryl orders beef, but not both.
c. Betty and Cheryl do not both order chicken.
Who could have ordered beef yesterday and chicken today?
19.
What are the last 3 digits of the sum 1 + 11 + 111 + … + 111…111?
70 digits
20.
The diagram shows 9 points. Draw 4 consecutive line segments (i.e. the start point of
the next segment must coincide with the endpoint of the previous segment) to pass
through all the 9 points.
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SASMO 2014 Round 1 Primary 4 Solutions
Section A [1 mark for each question]
1.
Find the missing term in the following sequence: 1, 4, 9, 16, _____, 36.
Solution
Observe that 1 = 12, 4 = 22, 9 = 32, 16 = 42 and 36 = 62.
 the missing term is 52 = 25.
2.
Find the smallest whole number between 1 and 100 that is divisible by 5 and by 7.
Solution
Smallest whole number between 1 and 100 that is divisible by 5 and by 7 = 5  7 = 35
3.
A frog fell into a drain that was 20 cm deep. After one hour, it mastered enough
energy to make a jump of 6 cm but it then slid down 4 cm. If it continued in this
manner after every one hour, how many hours will it take to get out of the drain?
Solution
After 7 hours, the frog would have jumped 7  2 = 14 cm.
In the next hour, it would have jumped the last 6 cm and out of the drain.
 it takes the frog 7 + 1 = 8 hours to get out of the drain.
4.
The diagram shows a circle with centre O and radius 5 cm. OABC is a rectangle. What
is the length of AC?
B
A
O
C
Solution
AC = OB (diagonal of rectangle)
= 5 cm (radius of circle)
5.
A textbook is opened at random. To what pages is it opened if the product of the
facing pages is 420?
Solution
Since 20  20 = 400, try 20  21 = 420.
 the pages are 20 and 21.
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6.
A shop sells sweets where every 3 sweet wrappers can be exchanged for one more
sweet. Kelvin has enough money to buy only 19 sweets. What is the biggest number
of sweets that he can get from the shop?
Solution
19 sweets  19 wrappers  6 sweets and 1 wrappers  7 wrappers  2 sweets and
1 wrapper  3 wrappers  1 sweet
 biggest no. of sweets = 19 + 6 + 2 + 1 = 28
7.
A clock takes 6 seconds to make 3 chimes. Assuming that the rate of chiming is
constant, and the duration of each chime is negligible, how long does the clock take to
make 4 chimes?
Solution
Time taken by the clock between 2 chimes = 6 s / 2 gaps = 3 s
 time taken by the clock to make 4 chimes = 3 s  3 gaps = 9 s
8.
The diagram shows a square being divided into four rectangles. If the sum of the
perimeter of the four rectangles is 32 cm, find the area of the square.
Solution
Perimeter of the four rectangles = 8  length of square = 32 cm
So length of square = 4 cm
Area of square = 16 cm2
9.
Alvin, Betty and Cheryl scored a total of 2014 points during a competition. Betty
scored 271 points less than Alvin. Betty scored 3 times as many points as Cheryl.
How many points did Betty score?
Solution
Method 1 (Model Method)
Cheryl
Betty
Alvin
2014
271
7 units = 2014  271 = 1743
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Method 2
There are 1 white square on the left side and 1 white square on the right side of the
row of black squares.
The pattern for the no. of white squares is 3  2 + 2, 4  2 + 2, 5  2   
 figure number with 2014 white squares = (2014  2)  2  2 = 1004
Method 3
The pattern for the no. of white squares is 3  3  1, 3  4  2, 3  5  3, … (total no.
of squares minus no. of black squares in each figure)
Thus no. of white squares that will surround one row of n black squares = 3(n + 2)  n
= 2n + 6
 figure number with 2014 white squares = (2014  6)  2 = 1004
Method 4
The pattern for the no. of white squares is 8, 10, 12, …, which is equal to 2  4, 2  5,
2  6, …
 figure number with 2014 white squares = 2014  2  3 = 1004
13.
Find the total number of squares in a 4  4 square grid.
Solution
No. of 1  1 squares = 16 = 42
No. of 2  2 squares = 9 = 32
No. of 3  3 squares = 4 = 22
No. of 4  4 squares = 1 = 12
 total no. of squares in a 4  4 square grid = 12 + 22 + 32 + 42 = 30
14.
How many digits are there before the fifteen 7 of the following number?
37337333733337333337…
Solution
No. of 7s before the fifteen 7 of the number = 14
No. of 3s before the fifteen 7 of the number = 1 + 2 + 3 + … + 15
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Method 1
1 + 15 = 16
2 + 14 = 16
3 + 13 = 16
7 pairs
7 + 9 = 16
8
 total no. of digits before the fifteen 7 of the number = 14 + (7  16) + 8 = 134
Method 2
Since 1 + 2 + 3 + … + n =
the number = 14 +
15.
, then total no. of digits before the hundredth 7 of
= 134
A box contains 30 coloured pens: 15 blue, 10 red and 5 black. Alice takes some pens
from the box without looking at the colours of the pens. What is the least number of
pens she must take so that she has at least 8 pens of the same colour?
Solution
Worst case scenario for all the pens taken out by Alice to be of different colours:
7 blue, 7 red and 5 black, i.e. a total of 19 pens.
Then the next pen taken out has to be either blue or red, i.e. Alice will have at least 8
pens of the same colour, either 8 blue or 8 red pens.
 least no. of pens Alice must take so that she has at least 8 pens of the same colour
= 19 + 1 = 20
Section B [2 marks for each question]
16.
In the following alphametic, all the different letters stand for different digits. Find the
four-digit sum PEEL.
+
P
A
M
L
A
P
E
E
L
Solution
Since the addition of two digits will give a maximum of 18, or a maximum of 19 if
there is a carryover (or renaming) of 1, this means that the maximum carryover is
1.
So L = 9 and there is a carryover of 1 for L to give 10, i.e. P = 1 and E = 0.
 the four-digit sum PEEL is 1009.
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17.
A teacher has a bag of sweets to treat her class. If she gave 4 sweets to each student,
then she would have 82 sweets left. If she gave 9 sweets to each student, then she
would have 2 sweets left. How many students and how many sweets are there?
Solution
If the teacher gave 4 sweets to each student, then she would have 82 sweets left.
From the 82 sweets left, if she gave 5 more sweets to each student so that each student
has 9 sweets, then she would have 2 left.
This means that she gave a total of 82  2 = 80 sweets from the 82 sweets left to the
students.
Since each student receives only one 5 more sweets (i.e. the 5th, 6th, 7th, 8th and 9th
sweets), then there are 80  5 = 16 students.
 there are 16  9 + 2 = 146 sweets.
18.
When Amy, Betty and Cheryl eat out, each orders either beef or chicken.
a. If Amy orders beef, Betty orders chicken.
b. Either Amy or Cheryl orders beef, but not both.
c. Betty and Cheryl do not both order chicken.
Who could have ordered beef yesterday and chicken today?
Solution
If Amy orders beef, Condition a says that Betty orders chicken, and Condition b says
that Cheryl also orders chicken, contradicting Condition c.
Beef
Chicken
Amy
(a)
Betty
Cheryl
(a)
(b)
Thus Amy always orders chicken.
Then Condition b says that Cheryl always orders beef.
Amy
Beef
Chicken
Betty
Cheryl
(b)

So the only person who can change her order is Betty.
 Betty ordered beef yesterday and chicken today.
19.
What are the last 3 digits of the sum 1 + 11 + 111 + … + 111…111?
70 digits
Solution
Last digit: 70  0 (and carry 7)
Second last digit: 69 + 7 (carry) = 76  6 (and carry 7)
Third last digit: 68 + 7 (carry) = 75  5
 the last 2 digits are 560.
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20.
The diagram shows 9 points. Draw 4 consecutive line segments (i.e. the start point of
the next segment must coincide with the endpoint of the previous segment) to pass
through all the 9 points.
Solution
If you try to draw the line segments within the region bounded by the dots, you will
realise that you need at least 5 consecutive line segments.
 you must draw some of the line segments outside the region as shown:
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SASMO 2014 Round 1 Primary 5 Problems
1.
If n is a whole number, for what values of n is
also a whole number?
2.
A textbook is opened at random. To what pages is it opened if the product of the
facing pages is 600?
3.
The diagram shows a quadrant OAB of a circle with centre O. OPQR is a rectangle.
Given that PR = 7 cm, find the length of OA.
B
Q
R
O
P
A
4.
Find an even number between 300 and 400 that is divisible by 5 and by 7.
5.
A shop sells sweets where every 3 sweet wrappers can be exchanged for one more
sweet. Navin has enough money to buy only 29 sweets. What is the biggest number of
sweets that he can get from the shop?
6.
Find the next term of the following sequence: 2, 3, 4, 10, 38, …
7.
The percentage passes in an exam for two classes are 80% and 60%. The numbers of
students in the two classes are 20 and 30 respectively. Find the overall percentage
pass for the two classes.
8.
A clock takes 9 seconds to make 4 chimes. Assuming that the rate of chiming is
constant, and the duration of each chime is negligible, how long does the clock take to
make 3 chimes?
9.
1 
 1  1  1  
Evaluate 1  1  1  1 
.
 2  3  4   2014 
10.
A frog fell into a drain that was 50 cm deep. After one hour, it mastered enough
energy to make a jump of 6 cm but it then slid down 4 cm. If it continued in this
manner after every one hour, how many hours will it take to get out of the drain?
11.
A farmer’s chickens produced 4028 eggs one day. Was he able to pack all the eggs in
full cartons of one dozen eggs each?
12.
A farmer wants to find out the number of sheep and ducks that he has. He counted a
total of 40 heads and 124 legs. How many sheep and how many ducks does he have?
13.
Jaime puts some blue and red cubes in a box. The ratio of the number of blue cubes to
the number of red cubes is 2 : 1. She adds 12 more red cubes in the box and the ratio
becomes 4 : 5. How many blue cubes are there in the box?
14.
The diagram shows a rectangle with its two diagonals. What percentage of the
rectangle is shaded?
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15.
Given that a  b = 2014, and a and b are whole numbers such that a < b, how many
possible pairs (a, b) are there?
16.
Amy had 3 times as much money as Betty. After they had spent $60 each, Amy had 4
times as much money as Betty. How much money did Amy have at first?
17.
Billy uses identical square tiles to make the following figures. If he continues using
the same pattern, in which figure will there be 6044 tiles?
Figure 1
Figure 2
Figure 3
Figure 4
18.
What is the least number of cuts required to cut 12 identical sausages so that they can
be shared equally among 20 people?
19.
In the following alphametic, all the different letters stand for different digits. Find the
two-digit sum PI.
+
I
S
I
S
I
S
I
S
P
I
20.
A teacher has a bag of sweets to treat her class. If she gave 6 sweets to each student,
then she would have 5 sweets left. If she gave 7 sweets to each student, then she
would have 30 sweets short. How many students and how many sweets are there?
21.
A box contains 80 coloured pens: 36 black, 24 blue, 12 red and 8 green. Alice takes
some pens from the box without looking at the colours of the pens. What is the least
number of pens she must take so that she has at least 20 pens of the same colour?
22.
Find the value of
23.
The diagram shows a square being divided into four rectangles. If the sum of the
perimeter of the four rectangles is 40 cm, find the area of the square.
24.
Given that 5! means 5  4  3  2  1, find the last digit of 2014!.
.
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25.
Two women, Ann and Carol, and two men, Bob and David, are athletes. One is a
swimmer, a second is a skater, a third is a gymnast, and a fourth is a tennis player. On
a day they were seated around a square table:
a. The swimmer sat on Ann’s left.
b. The gymnast sat across from Bob.
c. Carol and David sat next to each other.
d. A woman sat on the skater’s left.
Who is the tennis player?
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SASMO 2014 Round 1 Primary 5 Solutions
1.
If n is a whole number, for what values of n is
also a whole number?
Solution
is a whole number if n is a factor of 24.
24 = 1  24
= 2  12
=38
=46

2.
is a whole number if n = 1, 2, 3, 4, 6, 8, 12, 24.
A textbook is opened at random. To what pages is it opened if the product of the
facing pages is 600?
Solution
Since 25  25 = 625, try 24  25 = 600.
 the pages are 24 and 25.
3.
The diagram shows a quadrant OAB of a circle with centre O. OPQR is a rectangle.
Given that PR = 7 cm, find the length of OA.
B
Q
R
O
P
A
Solution
OA = OQ (radii of quadrant)
= PR (diagonal of rectangle)
= 7 cm
4.
Find an even number between 300 and 400 that is divisible by 5 and by 7.
Solution
Since 2, 5 and 7 are relatively prime, if a number is divisible by 2, by 5 and by 7, then
the number is divisible by 2  5  7 = 70, i.e. the number is a multiple of 70.
 an even number between 300 and 400 that is divisible by 5 and by 7 is 350.
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5.
A shop sells sweets where every 3 sweet wrappers can be exchanged for one more
sweet. Navin has enough money to buy only 29 sweets. What is the biggest number of
sweets that he can get from the shop?
Solution
29 sweets  29 wrappers  9 sweets and 2 wrappers  11 wrappers  3 sweets
and 2 wrappers  5 wrappers  1 sweet and 2 wrappers  3 wrappers  1 sweet
 biggest no. of sweets = 29 + 9 + 3 + 1 + 1 = 43
6.
Find the next term of the following sequence: 2, 3, 4, 10, 38, …
Solution
From the third term onwards, the next term is obtained by multiplying the previous
two terms and then subtracting 2.
 the next term is 10  38  2 = 378.
7.
The percentage passes in an exam for two classes are 80% and 60%. The numbers of
students in the two classes are 20 and 30 respectively. Find the overall percentage
pass for the two classes.
Solution
No. of students in the first class who pass the exam = 80%  20 = 16
No. of students in the second class who pass the exam = 60%  30 = 18
Total no. of students in both classes who pass the exam = 16 + 18 = 34
 overall percentage pass for the two classes = 34 / 50  100% = 68%
Note: Common mistake is 70%. But class size is different: the weaker class will pull
down the percentage because it has more students than the better class. So you
can’t average percentages unless the base (in this case, the number of students
in each class) is the same.
8.
A clock takes 9 seconds to make 4 chimes. Assuming that the rate of chiming is
constant, and the duration of each chime is negligible, how long does the clock take to
make 3 chimes?
Solution
Time taken by the clock between 2 chimes = 9 s / 3 gaps = 3 s
 time taken by the clock to make 3 chimes = 3 s  2 gaps = 6 s
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9.
1 
 1  1  1  
Evaluate 1  1  1  1 
.
 2  3  4   2014 
Solution
1   1  2  3   2013 
 1  1  1  
1  1  1  1 
 =     

 2  3  4   2014   2  3  4   2014 
1
=
2014
10.
A frog fell into a drain that was 50 cm deep. After one hour, it mastered enough
energy to make a jump of 6 cm but it then slid down 4 cm. If it continued in this
manner after every one hour, how many hours will it take to get out of the drain?
Solution
After 22 hours, the frog would have jumped 22  2 = 44 cm.
In the next hour, it would have jumped the last 6 cm and out of the drain.
 it takes the frog 22 + 1 = 23 hours to get out of the drain.
11.
A farmer’s chickens produced 4028 eggs one day. Was he able to pack all the eggs in
full cartons of one dozen eggs each?
Solution
A dozen is equal to 12.
If a number is divisible by 12 (= 3  4), then it is also divisible by 3 and 4, since 3 and
4 are relatively prime.
Using the divisibility test for 4, 4028 is divisible by 4 since the last two digits 28 is
divisible by 4.
Using the divisibility test for 3, 4028 is not divisible by 3 since 4 + 0 + 2 + 8 = 14 is
not divisible by 3, i.e. 4028 is not divisible by 12.
 the farmer was not able to pack all the eggs in full cartons of one dozen eggs.
12.
A farmer wants to find out the number of sheep and ducks that he has. He counted a
total of 40 heads and 124 legs. How many sheep and how many ducks does he have?
Solution
Method 1 (Supposition)
Suppose all the animals are ducks.
Then there would be 40  2 = 80 legs.
So the remaining 124  80 = 44 legs must belong to the “extra 2 legs” of the sheep.
 there are 44  2 = 22 sheep and 40  22 = 18 ducks.
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Method 2 (Systematic Guess and Check)
Make a systematic list, starting with a random guess: 20 sheep and 20 ducks.
No. of Sheep
No. of Ducks
20
21
22
20
19
18
No. of Sheep
Legs
80
84
88
No. of Duck
Legs
40
38
36
Total No. of
Legs
120
122
124
 there are 22 sheep and 18 ducks.
Method 3 (Algebra)
Let the no. of sheep be x.
Then there are 40  x ducks.
Total no. of legs = 4x + 2(40  x) = 124
4x + 80  2x = 124
2x = 44
x = 22
 there are 22 sheep and 40  22 = 18 ducks.
13.
Jaime puts some blue and red cubes in a box. The ratio of the number of blue cubes to
the number of red cubes is 2 : 1. She adds 12 more red cubes in the box and the ratio
becomes 4 : 5. How many blue cubes are there in the box?
Solution
Method 1 (Model Method)
Before
Blue
Red
After
Blue
Red
12
From the diagram, 3 units = 12
1 unit = 4
4 units = 16
 there are 16 blue cubes.
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Method 2 (Algebraic Method)
Let the no. of red cubes at the start be x.
Then the no. of blue cubes will be 2x.
After adding 12 more red cubes, there are x + 12 red cubes.
Ratio of blue cubes to red cubes is now 4 : 5, i.e.
.
Then 10x = 4x + 48
6x = 48
x=8
2x = 16
 there are 16 blue cubes.
14.
The diagram shows a rectangle with its two diagonals. What percentage of the
rectangle is shaded?
Solution
Let the length and breadth of a rectangle be l and b respectively.
Then area of rectangle = lb
and area of shaded triangle =  base  height =  b  = .
 percentage of the rectangle that is shaded =  100% = 25%
15.
Given that a  b = 2014, and a and b are whole numbers such that a < b, how many
possible pairs (a, b) are there?
Solution
2014 = 2  19  53, where 2, 19 and 53 are prime numbers.
2014 = 1  2014
= 2  1007
= 19  106
= 38  53
 there are 4 possible pairs (a, b).
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16.
Amy had 3 times as much money as Betty. After they had spent $60 each, Amy had 4
times as much money as Betty. How much money did Amy have at first?
Solution
Method 1 (Model Method)
After
Amy
Betty
Before
$60
Amy
Betty
$60
$60
$60
From the Before diagram, where Amy had 3 times as much money as Betty,
1 unit = $60 + $60 = $120
4 units = $480
 Amy had $480 + $60 = $540 at first.
Method 2 (Algebraic Method)
Let the amount of money Betty had at first be $x.
Then the amount of money Alice had at first was $3x.
After they had spent $60 each, Betty had $(x  60) and Alice had $(3x  60).
Then $(3x  60) = 4  $(x  60)
3x  60 = 4x  240
x = 180
3x = $540
 Amy had $540 at first.
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17.
Billy uses identical square tiles to make the following figures. If he continues using
the same pattern, in which figure will there be 6044 tiles?
Figure 1
Figure 2
Figure 3
Figure 4
Solution
Method 1
The two corner tiles are the same for all figures.
 figure number with 6044 tiles = (6044  2)  3 = 2014
Method 2
The tiles in the top row have this pattern: 3, 4, 5, 6, …
Thus the nth figure will have (n + 2) + 2  n = 3n + 2 tiles.
 figure number with 6044 tiles = (6044  2)  3 = 2014
Method 3
The tiles in each of the vertical column have this pattern: 2, 3, 4, 5, …
Thus the nth figure will have (n + 1)  2 + n = 3n + 2 tiles.
 figure number with 6044 tiles = (6044  2)  3 = 2014
Method 4
The no. of tiles in each figure is equal to the “area of the rectangle” minus the “area of
the hole in the middle”.
The pattern for the “area of the rectangle” is 2  3, 3  4, 4  5, 5  6, …
The pattern for the “area of the hole in the middle” is 1  1, 2  2, 3  3, 4  4, …
Thus the nth figure will have (n + 1)  (n + 2)  n  n = 3n + 2 tiles.
 figure number with 6044 tiles = (6044  2)  3 = 2014
18.
What is the least number of cuts required to cut 12 identical sausages so that they can
be shared equally among 20 people?
Solution
Fraction of sausage each person will get =
=
This means that there must be at least 12 cuts since no one will get one whole
sausage.
Cut each of the 12 sausages at the -mark. Then 12 people will get one
sausage
each.
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There are 12 times sausages left. But each person must get of a sausage.
This means that 4 of the
sausages must be cut into half each so that the remaining 8
people will get one sausage and one sausage each.
 least no. of cuts = 12 + 4 = 16
19.
In the following alphametic, all the different letters stand for different digits. Find the
two-digit sum PI.
+
I
S
I
S
I
S
I
S
P
I
Solution
If I  3, then the sum will be a 3-digit number.
So I = 1 or 2.
In the ones column, S + S + S + S = 4  S = _I.
Since 4  S is even, then I must be even, so I = 2.
But 4  S = _2 implies that S = 3 or 8.
If S = 8, then 28  4 is a 3-digit number, so S = 3.
Thus 23 + 23 + 23 + 23 = 92, i.e. P = 9.
 the two-digit sum PI is 91.
20.
A teacher has a bag of sweets to treat her class. If she gave 6 sweets to each student,
then she would have 5 sweets left. If she gave 7 sweets to each student, then she
would have 30 sweets short. How many students and how many sweets are there?
Solution
If the teacher gave 6 sweets to each student, then she would have 5 sweets left.
But 5 sweets left are not enough to give one more sweet to each student since she
would have 30 sweets short.
So if she had 30 more sweets, then she could give 30 + 5 = 35 sweets to the students
so that each student has one more sweet, i.e. 6 sweets each.
Since each student receives only 1 more sweet (i.e. the 6th sweet), then there are 35
students.
 there are 35  6 + 5 = 215 sweets.
21.
A box contains 80 coloured pens: 36 black, 24 blue, 12 red and 8 green. Alice takes
some pens from the box without looking at the colours of the pens. What is the least
number of pens she must take so that she has at least 20 pens of the same colour?
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Solution
Worst case scenario for all the pens taken out by Alice to be of different colours:
19 black, 19 blue, 12 red and 8 green, i.e. a total of 58 pens.
Then the next pen taken out has to be either black or blue, i.e. Alice will have at least
20 pens of the same colour, either 20 black or 20 blue pens.
 least no. of pens Alice must take so that she has at least 20 pens of the same colour
= 58 + 1 = 59
22.
Find the value of
.
Solution
Observe the following pattern:

23.
.
The diagram shows a square being divided into four rectangles. If the sum of the
perimeter of the four rectangles is 40 cm, find the area of the square.
Solution
Perimeter of the four rectangles = 8  length of square = 40 cm
So length of square = 5 cm
Area of square = 25 cm2
24.
Given that 5! means 5  4  3  2  1, find the last digit of 2014!.
Solution
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Since 2014! contains the factor 10, and any number multiplied by 10 will have 0 as
the last digit, then the last digit of 2014! is 0.
25.
Two women, Ann and Carol, and two men, Bob and David, are athletes. One is a
swimmer, a second is a skater, a third is a gymnast, and a fourth is a tennis player. On
a day they were seated around a square table:
a. The swimmer sat on Ann’s left.
b. The gymnast sat across from Bob.
c. Carol and David sat next to each other.
d. A woman sat on the skater’s left.
Who is the tennis player?
Solution
There are only two possibilities to satisfy the first two conditions:
Bob
Bob
swimmer
swimmer
gymnast
Ann
gymnast
Ann
Possibility 1
Possibility 2
Only Possibility 2 satisfies Condition c.
Condition d says that Ann cannot be the skater since Bob is on her left.
 Ann must be the tennis player.
10
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SASMO 2014 Round 1 Primary 6 Problems
1.
A clock takes 3 seconds to make 4 chimes. Assuming that the rate of chiming is
constant, and the duration of each chime is negligible, how long does the clock take to
make 8 chimes?
2.
If n is a whole number, for what values of n is
3.
Find the smallest whole number between 1 and 100 that is divisible by 12 and by 30.
4.
A shop sells sweets where every 3 sweet wrappers can be exchanged for one more
sweet. Catherine has enough money to buy only 26 sweets. What is the biggest
number of sweets that she can get from the shop?
8.
A whole number is between 50 and 100. When it is divided by 5, the remainder is 3.
When it is divided by 7, the remainder is 5. Find the number.
9.
A man travelled at 120 km/h for the first half of a 12-km journey. Then he travelled at
60 km/h for the rest of his journey. What is his average speed for the whole journey?
10.
Amy buys an item for a 20% discount during a sale, but she still needs to pay a 5%
GST (Goods and Services Tax). She is given two options.
also a whole number?
Option A: 20% discount first, then add 5% GST
Option B: Add 5% GST first, then 20% discount
Which option is cheaper for Amy? Or does it not matter?
11.
The digits 3, 4, 5, 6 and 7 are all used to write a five-digit number ABCDE where the
three-digit number ABC is divisible by 4, BCD is divisible by 5 and CDE is divisible
by 3. Find all the possible values of the five-digit number ABCDE.
12.
At a workshop, there are 20 participants. Each of them shakes hand once with one
another. How many handshakes are there?
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13.
In the following cryptarithm, different letters represent different digits and an asterisk
* may represent any digit. The product MATH is a four-digit number less than 5000.
What number does MATH represent?
E
H
E
H
*
*
6
*
8
M A
T

+
*
H
14.
There are 100 buns to be shared among 100 monks (consisting of senior and junior
monks). The senior monks get 3 buns each and every 3 junior monks share 1 bun.
How many senior monks are there?
15.
What is the least number of cuts required to cut 5 identical sausages so that they can
be shared equally among 9 people?
16.
How many digits are there before the 50th ‘8’ of the following number?
85855855585555855555…
17.
Find the total number of triangles in the diagram.
18.
A teacher has a bag of sweets to treat her class. If she gave 5 sweets to each student,
then she would have 13 sweets left. If she gave 8 sweets to each student, then she
would have 20 sweets short. How many students and how many sweets are there?
19.
Find the value of
20.
Susie has 23 coins. She divides them into 5 piles so that each pile has a different
number of coins. Find the smallest possible number of coins in the biggest pile.
21.
Find the next term of the following sequence: 1, 4, 9, 7, 7, 9, 4, 1, 9, 1, …
22.
Given that 5! means 5  4  3  2  1, find the last two digits of 2014!.
23.
There were 210 students in the hall. 2 of the boys and 1 of the girls wore T-shirts. If
.
5
3
there were 78 students in the hall who wore T-shirts, how many boys were in the hall?
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24.
A farmer wants to plant 10 trees in 5 rows such that there are exactly 4 trees in each
row. Draw a diagram to show how the trees should be planted.
25.
Frank knows 5 women: Amy, Betty, Cheryl, Doris and Elaine.
a. 3 women are under 30 and the other 2 women are over 30.
b. 3 women are nurses and the other 2 women are teachers.
c. Amy and Cheryl are in the same age bracket.
d. Doris and Elaine are in different age brackets.
e. Betty and Elaine have the same occupation.
f. Cheryl and Doris have different occupations.
g. Of the 5 women, Frank will marry the teacher over 30.
Who will Frank marry?
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SASMO 2014 Round 1 Primary 6 Solutions
1.
A clock takes 3 seconds to make 4 chimes. Assuming that the rate of chiming is
constant, and the duration of each chime is negligible, how long does the clock take to
make 8 chimes?
Solution
Time taken by the clock between 2 chimes = 3 s / 3 gaps = 1 s
 time taken by the clock to make 8 chimes = 1 s  7 gaps = 7 s
2.
If n is a whole number, for what values of n is
also a whole number?
Solution
is a whole number if n is a factor of 36.
36 = 1  36
= 2  18
= 3  12
=49
=66

3.
is a whole number if n = 1, 2, 3, 4, 6, 9, 12, 18, 36.
Find the smallest whole number between 1 and 100 that is divisible by 12 and by 30.
Solution
Smallest whole number between 1 and 100 that is divisible by 12 and by 30
= LCM(12, 30) = 60
4.
A shop sells sweets where every 3 sweet wrappers can be exchanged for one more
sweet. Catherine has enough money to buy only 26 sweets. What is the biggest
number of sweets that she can get from the shop?
Solution
Method 1
26 sweets  26 wrappers  8 sweets and 2 wrappers  10 wrappers  3 sweets
and 1 wrapper  4 wrappers  1 sweet and 2 wrappers
 biggest no. of sweets = 26 + 8 + 3 + 1 = 38 [Common mistakes: 34, 37]
Method 2
26 sweets  8 sweets and 2 wrappers  2 sweets and 4 wrappers  2 sweets
 biggest no. of sweets = 26 + 8 + 2 + 2 = 38 [Common mistakes: 34, 36]
1
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Method 2 (Model Method)
4 kg
Duck
1.2
Chicken
The idea of average is to “even out”.
This means that the 4-kg mark will divide 1.2 kg into 2 equal parts.
 mass of duck = 4 kg + 0.6 kg = 4.6 kg
Method 3 (Algebra)
Let the mass of the chicken be x kg.
Then the mass of the duck is (x + 1.2) kg.
Average mass of the duck and the chicken =
=4
2x + 1.2 = 8
2x = 6.8
x = 3.4
 mass of duck = 3.4 + 1.2 = 4.6 kg
8.
A whole number is between 50 and 100. When it is divided by 5, the remainder is 3.
When it is divided by 7, the remainder is 5. Find the number.
Solution
Since the number leaves a remainder of 3 when divided by 5, then it must end with 3
or 8, i.e. the possible values of the number are 53, 58, 63, 68, …, 98.
Since the number leaves a remainder of 3 when divided by 5, we can either check
each of the above possible values one by one, or we can start with 7  7 + 5 = 54, and
then add 7 until we get a number ending in 3 or 8, i.e. 54, 61, 68, …
 the number is 68.
9.
A man travelled at 120 km/h for the first half of a 12-km journey. Then he travelled at
60 km/h for the rest of his journey. What is his average speed for the whole journey?
Solution
Time taken for first half of the 12-km journey = 6 km / 120 km/h = 0.05 h
Time taken for second half of the 12-km journey = 6 km / 60 km/h = 0.1 h
 average speed for whole journey = 12 km / 0.15 h = 80 km/h
Note: Common mistake is 90 km/h. Since speed = distance / time, if you want to
average speeds, then the base (in this case, the time taken) must be the same.
But instead the distance travelled for both parts of the journey is the same, so
you cannot average speeds.
3
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10.
Amy buys an item for a 20% discount during a sale, but she still needs to pay a 5%
GST (Goods and Services Tax). She is given two options.
Option A: 20% discount first, then add 5% GST
Option B: Add 5% GST first, then 20% discount
Which option is cheaper for Amy? Or does it not matter?
Solution
Option A: 0.8  1.05  Sales Price
Option B: 1.05  0.8  Sales Price
 both options are the same, so it does not matter.
Note: Common mistake is Option A. If you pay 5% GST first (Option B), although
this GST is bigger than the GST in Option A, this GST is also discounted 20%,
so it ends up the same as if you pay the GST after the discount (Option A).
Notice that the base is the same (in this case, the selling price).
11.
The digits 3, 4, 5, 6 and 7 are all used to write a five-digit number ABCDE where the
three-digit number ABC is divisible by 4, BCD is divisible by 5 and CDE is divisible
by 3. Find all the possible values of the five-digit number ABCDE.
Solution
Since BCD divisible by 5, then D = 5.
Since ABC divisible by 4, then BC is also divisible by 4, i.e. BC = 36 or 64.
Since CDE is divisible by 3, then C + D + E = C + 5 + E is also divisible by 3.
If BC = 36, then C + 5 + E = 6 + 5 + E = 11 + E is also divisible by 3, i.e. E = 4 or 7,
so A = 7 or 4 respectively.
If BC = 64, then C + 5 + E = 4 + 5 + E = 9 + E is also divisible by 3, i.e. E = 3, so A =
7.
 the possible values of ABCDE are 43657, 73654 and 76453.
12.
At a workshop, there are 20 participants. Each of them shakes hand once with one
another. How many handshakes are there?
Solution
The first participant will shake hand with 19 other participants;
the second participant will shake hand with 18 other participants;
the third participant will shake hand with 17 other participants; etc.
Thus total no. of handshakes = 19 + 18 + 17 + … + 3 + 2 + 1
Method 1
1 + 19 = 20
2 + 18 = 20
3 + 17 = 20
9 pairs
4
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9 + 11 = 20
10
 total no. of handshakes = 20  9 + 10 = 190
Method 2
Since 1 + 2 + 3 + … + n =
13.
, then total no. of handshakes =
= 190
In the following cryptarithm, different letters represent different digits and an asterisk
* may represent any digit. The product MATH is a four-digit number less than 5000.
What number does MATH represent?
E
H
E
H
*
*
6
*
8
M A
T

+
*
H
Solution
In the ones column, H  H = _6 implies H = 6.
In the tens column, E  H = E  6 = _8 implies E = 3 or 8.
If E = 8, then 86  86 > 5000. So E = 3.
 MATH = 36  36 = 1296.
14.
There are 100 buns to be shared among 100 monks (consisting of senior and junior
monks). The senior monks get 3 buns each and every 3 junior monks share 1 bun.
How many senior monks are there?
Solution
Method 1 (By Observation)
Observe that if you group one senior monk and 3 junior monks together, they will get
a total of 4 buns.
And 25 such groups will give a total of 100 monks and 100 buns.
 there are 25 senior monks.
Method 2 (Algebra)
Let the no. of senior monks be x.
Then the no. of junior monks is 100  x.
Total no. of buns = 3x +
= 100
9x + (100  x) = 300
5
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8x = 200
x = 25
 there are 25 senior monks.
15.
What is the least number of cuts required to cut 5 identical sausages so that they can
be shared equally among 9 people?
Solution
Fraction of sausage each person will get =
This means that there must be at least 5 cuts since no one will get one whole sausage.
Cut each of the 5 sausages at the -mark. Then 5 people will get one sausage each.
There are 5 times sausages left. But each person must get of a sausage.
This means that one of the
sausages must be cut into 4 equal parts each (i.e. 3 cuts
each) so that the remaining 4 people will get one sausage and one sausage each.
 least no. of cuts = 5 + 3 = 8
16.
How many digits are there before the 50th ‘8’ of the following number?
85855855585555855555…
Solution
No. of ‘8’s before the 50th ‘8’ of the number = 49
No. of ‘5’s before the 50th ‘8’ of the number = 1 + 2 + 3 + … + 49
Method 1
1 + 49 = 50
2 + 48 = 50
3 + 47 = 50
24 pairs
24 + 26 = 50
25
 total no. of digits before the 50th ‘8’ = 49 + 50  24 + 25 = 1274
Method 2
Since 1 + 2 + 3 + … + n =
number = 49 +
, then total no. of digits before the 50th ‘8’ of the
= 1274
6
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17.
Find the total number of triangles in the diagram.
Solution
1
Label the vertices as shown in the diagram.
These are the vertices that form a triangle:
123, 124, 125, 126, 134, 135, 137, 145, 156, 157,
158, 167, 168, 236, 256, 347, 356, 357, 367, 457,
458, 568, 578, 678
 there is a total of 24 triangles in the diagram.
18.
3
2
4
5
7
8
6
A teacher has a bag of sweets to treat her class. If she gave 5 sweets to each student,
then she would have 13 sweets left. If she gave 8 sweets to each student, then she
would have 20 sweets short. How many students and how many sweets are there?
Solution
If the teacher gave 5 sweets to each student, then she would have 13 sweets left.
But 13 sweets left are not enough to give 3 more sweets to each student since she
would have 20 sweets short.
So if she had 20 more sweets, then she could give 20 + 13 = 33 sweets to the students
so that each student has 3 more sweets, i.e. 8 sweets each.
Since each student receives only 3 more sweets (i.e. the 6th, 7th and 8th sweets), then
there are 33  3 = 11 students.
 there are 11  5 + 13 = 68 sweets.
19.
Find the value of
.
Solution
Method 1
Observe the following pattern:

.
7
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Method 2
Since
20.
for all n > 1, then
Susie has 23 coins. She divides them into 5 piles so that each pile has a different
number of coins. Find the smallest possible number of coins in the biggest pile.
Solution
For each pile to have a different number of coins, and the biggest pile to have the
smallest possible number of coins, put 1 coin in the 1st pile, 2 coins in the 2nd pile,
3 coins in the 3rd pile, 4 coins in the 4th pile and 5 coins in the 5th pile. So the
biggest pile is the 5th pile, but there are only 1 + 2 + 3 + 4 + 5 = 15 coins.
The 16th coin will have to go to the 5th pile so that each pile will have a different
number of coins. The 17th coin cannot go to the 5th pile because we want to find the
smallest possible number of coins in the biggest pile, so the 17th coin will have to
go to the 4th pile. Similarly, the 18th, 19th and 20th coins will go to the 3rd, 2nd and
1st piles respectively.
The 21st coin will go to the 5th pile again, the 22nd coin to the 4th pile, and the 23rd coin
to the 3rd pile.
 the largest pile (which is the 5th pile) will contain 5 + 1 + 1 = 7 coins.
21.
Find the next term of the following sequence: 1, 4, 9, 7, 7, 9, 4, 1, 9, 1, …
Solution
Observe that the first 3 terms are perfect squares: 12, 22 and 32. But subsequent terms
are not perfect squares.
However, if you compare the given sequence with perfect squares (1, 4, 9, 16, 25, 36,
49, 64, 81, 100, …), you will observe that the n-th term can be found by squaring n
and then adding the digits continuously until a single-digit number is obtained.
 the next term, which is the 11th term, is obtained by: 112 = 121  1 + 2 + 1 = 4.
22.
Given that 5! means 5  4  3  2  1, find the last two digits of 2014!.
Solution
Since 2014! contains the factor 100, and any number multiplied by 100 will have 00
as the last two digits, then the last two digits of 2014! are 00.
8
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23.
There were 210 students in the hall. 2 of the boys and 1 of the girls wore T-shirts. If
5
3
there were 78 students in the hall who wore T-shirts, how many boys were in the hall?
Solution
Method 1 (Model Method)
The first model shows 2 of the boys and 1 of the girls wearing T-shirts (indicated by
5
3
solid boxes, not the dotted ones). We will call an unshaded box from the boys as Bbox, and a shaded box from the girls as G-box.
Boys
B1
B1
B2
B2
210
Girls
G1
G2
The 2 solid B-boxes and the one solid G-box (i.e. B1 + B1 + G1) represent a total of 78
students.
If you take 2 dotted B-boxes and 1 dotted G-box (i.e. B2 + B2 + G2), the 3 dotted
boxes also represent a total of 78 students.
But you can’t do anything with the remaining 2 dotted boxes.
So we double all the boxes as shown in the second model, where the total no. of
students is 210  2 = 420.
Boys
B1
Girls
G1
B1
B2
B2
B3
B3
B4
B4
B5
B5
420
G2
G3
G4
G5
From the second model, we are able to form 5 groups of 2 B-boxes and 1 G-box (i.e.
from B1 and G1 to B5 and G5), leaving behind only one G-box, where each group of
2 B-boxes and 1 G-box represent a total of 78 students.
This means that the last G-box represents 420  5  78 = 420  390 = 30 students.
 there are 30  3 = 90 girls and 210  90 = 120 boys in the hall.
Method 2 (Algebra)
Let the no. of boys be x.
Then the no. of girls is 210  x.
Total no. of students who wore T-shirts = 2 x + 1 (210  x) = 78
5
3
6x + 5(210  x) = 78  15
6x + 1050  5x = 1170
x = 120
 there are 120 boys.
9
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24.
A farmer wants to plant 10 trees in 5 rows such that there are exactly 4 trees in each
row. Draw a diagram to show how the trees should be planted.
Solution
If you start with 10 trees and then try fitting 5 lines onto them, you will realise that the
lines must overlap because there are not enough trees, and it’s not easy to fit 5 lines
onto 10 trees.
 try to draw 5 overlapping lines first, and a common figure with 5 overlapping lines
is the following star:
Then put in the trees and yes, it works.
25.
Frank knows 5 women: Amy, Betty, Cheryl, Doris and Elaine.
a. 3 women are under 30 and the other 2 women are over 30.
b. 3 women are nurses and the other 2 women are teachers.
c. Amy and Cheryl are in the same age bracket.
d. Doris and Elaine are in different age brackets.
e. Betty and Elaine have the same occupation.
f. Cheryl and Doris have different occupations.
g. Of the 5 women, Frank will marry the teacher over 30.
Who will Frank marry?
Solution
Conditions a, c and d  Amy and Cheryl are under 30, and Betty is over 30.
Conditions b, e and f  Betty and Elaine are nurses, and Amy is a teacher.
3 women under 30
Amy
Cheryl
?
2 women over 30
Betty
?
3 Nurses
Betty
Elaine
?
2 Teachers
Amy
?
So Amy is a teacher under 30.
Conditions b and g  the other teacher must be over 30.
Condition f  Cheryl or Doris is the other teacher over 30.
But Cheryl is under 30, so Doris is the other teacher over 30.
 Frank will marry Doris.
10
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SASMO 2014 Round 1 Secondary 1 Problems
1.
Find the next term of the following sequence: 2, 1, 3, 4, 7, …
2.
Find the product of the highest common factor and the lowest common multiple of 8
and 12.
3.
Solve for x and y in the following equation
4.
The last day of 2013 was a Tuesday. There are 365 days in 2014. In what day of the
week will 2014 end?
5.
What is the maximum number of parts that can be obtained from cutting a circular disc
using 3 straight cuts?
6.
A man bought two paintings and then sold them for $300 each. He made a profit of
20% for the first painting, but a loss of 20% for the second painting. Overall, did he
make a profit, a loss or break even? If he did not break even, state the amount of profit
or loss.
7.
Solve
8.
Given that xyz = 2014, and x, y and z are positive integers such that x < y < z, how
many possible triples (x, y, z) are there?
9.
At a workshop, there are 27 participants. Each of them shakes hand once with one
another. How many handshakes are there?
10.
A perfect number is a positive integer that is equal to the sum of its proper positive
factors. Proper positive factors of a number are positive factors that are less than the
number. For example, 6 = 1 + 2 + 3 is a perfect number because 1, 2 and 3 are the only
proper positive factors of 6. Find the next perfect number.
11.
The dimensions of a rectangle are x cm by y cm, where x and y are integers, such that
the area and perimeter of the rectangle are numerically equal. Find all the possible
values of x and y.
12.
If a and b are positive integers such that a < b and ab = ba, find a possible value for a
and for b.
13.
Find the value of
.
= 2.
1

1
1  2
1 2 1
1

 




.
14.
Find the last digit of 20142014.
15.
The diagram shows 9 points. Draw 4 consecutive line segments (i.e. the start point of
the next segment must coincide with the endpoint of the previous segment) to pass
through all the 9 points.
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16.
What are the last 5 digits of the sum 1 + 11 + 111 + … + 111…111?
2014 digits
17.
What is the least number of cuts required to cut 10 identical sausages so that they can
be shared equally among 18 people?
18.
Divide the following shape into 4 identical parts.
19.
Solve the following equation: x5 + 2x3  x2  2 = 0.
20.
Find the value of
21.
In the following cryptarithm, all the letters stand for different digits. Find the values of
A, B, C and D.

22.
.
A
8
B
C
3
D
9
8
2
0
1
4
Find the sum of the terms in the nth pair of brackets:
(1, 2), (3, 4), (5, 6), (7, 8), …
23.
In the diagram, PQ is parallel to RS, PA = PB and RB = RC. Given that BCA = 60,
find BAC.
A
P
Q
B
60
R
C
S
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24.
Find the remainder when 22014 is divided by 7.
25.
The diagram shows a triangle ABC where AB = AC, BC = AD and BAC = 20. Find
ADB.
A
20
D
B
C
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SASMO 2014 Round 1 Secondary 1 Solutions
1.
Find the next term of the following sequence: 2, 1, 3, 4, 7, …
Solution
From the third term onwards, the next term is obtained by adding the previous two
terms.
 the next term is 4 + 7 = 11.
Note: These are called Lucas numbers, which is a particular Lucas sequence.
2.
Find the product of the highest common factor and the lowest common multiple of 8
and 12.
Solution
Method 1
For any two natural numbers a and b, HCF(a, b)  LCM(a, b) = ab. [It does not work
for 3 or more numbers.]
 HCF(8, 12)  LCM(8, 12) = 8  12 = 96
Method 2
8 = 23
12 = 22  3
HCF(8, 12) = 22 = 4
LCM(8, 12) = 23  3 = 24
 HCF(8, 12)  LCM(8, 12) = 4  24 = 96
3.
Solve for x and y in the following equation
.
Solution
Since
and
way for
0.
 x = 7 and y = 8
4.
cannot be negative for any values of x and y, then the only
to be equal to 0 is when
= 0 and
=
The last day of 2013 was a Tuesday. There are 365 days in 2014. In what day of the
week will 2014 end?
Solution
365 days = 52 weeks and 1 day
Since the last day of 2013 was a Tuesday, then the last day of 2014 will be
Wednesday.
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5.
What is the maximum number of parts that can be obtained from cutting a circular disc
using 3 straight cuts?
Solution
Maximum number of parts = 7
6.
A man bought two paintings and then sold them for $300 each. He made a profit of
20% for the first painting, but a loss of 20% for the second painting. Overall, did he
make a profit, a loss or break even? If he did not break even, state the amount of profit
or loss.
Solution
Profit of 20% for first painting = $300 / 120  20 = $50.
Loss of 20% for second painting = $300 / 80  20 = $75.
 the man lost $25.
7.
Solve
= 2.
Solution
=2
=4
x+2=4
x=2
8.
Given that xyz = 2014, and x, y and z are positive integers such that x < y < z, how
many possible triples (x, y, z) are there?
Solution
2014 = 2  19  53, where 2, 19 and 53 are prime numbers.
 2014 = 1  1  2014
= 1  2  1007
= 1  19  106
= 1  38  53
= 2  19  53
 there are 5 possible pairs (x, y, z).
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9.
At a workshop, there are 27 participants. Each of them shakes hand once with one
another. How many handshakes are there?
Solution
The first participant will shake hand with 26 other participants;
the second participant will shake hand with 25 other participants;
the third participant will shake hand with 24 other participants; etc.
Thus total no. of handshakes = 26 + 25 + 24 + … + 3 + 2 + 1
Method 1
1 + 26 = 27
2 + 25 = 27
3 + 24 = 27
13 pairs
13 + 14 = 27
 total no. of handshakes = 27  13 = 351
Method 2
Since 1 + 2 + 3 + … + n =
10.
, then total no. of handshakes =
= 351
A perfect number is a positive integer that is equal to the sum of its proper positive
factors. Proper positive factors of a number are positive factors that are less than the
number. For example, 6 = 1 + 2 + 3 is a perfect number because 1, 2 and 3 are the only
proper positive factors of 6. Find the next perfect number.
Solution
By systematic trial and error from 7 to 28, since 28 = 1 + 2 + 4 + 7 + 14, the next
perfect number is 28.
11.
The dimensions of a rectangle are x cm by y cm, where x and y are integers, such that
the area and perimeter of the rectangle are numerically equal. Find all the possible
values of x and y.
Solution
Method 1
Area = perimeter
xy  2x  y 
2x
y
x2
Since x  2 > 0, then x > 2.
When x = 3, y = 6.
When x = 4, y = 4.
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When x = 5, y = is not an integer.
When x = 6, y = 3.
If x > 6, y < 3.
2y
By symmetry, x 
, i.e. y > 2.
y2
So there are no other solutions.
 the dimensions of all the rectangles with integral sides whose area and perimeter are
numerically equal are 3 by 6, 4 by 4, and 6 by 3.
Method 2
Area = perimeter
xy  2x  y 
2x
y
x2
2x  4  4

x2
2x  2  4

x2
4
 2
x2
Since x  2 > 0, then x > 2.
4
For y to be an integer,
must also be an integer.
x2
This means that 4  x – 2, i.e. x  6.
So the only possible solutions are when x = 3, 4, 5 and 6.
When x = 3, y = 6.
When x = 4, y = 4.
When x = 5, y = is not an integer.
When x = 6, y = 3.
 the dimensions of all the rectangles with integral sides whose area and perimeter are
numerically equal are 3 by 6, 4 by 4, and 6 by 3.
12.
If a and b are positive integers such that a < b and ab = ba, find a possible value for a
and for b.
Solution
By guess and check, a = 2, b = 4.
In fact, this is the only solution.
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13.
Find the value of
1

1
1  2
1 2 1
1

 




.
Solution
Let
1
= x. Then
1
= x, i.e. 2x2 + x  1 = 0.
1  2x


1

1  2
1 2 1 
1


 (2x  1) (x + 1) = 0, i.e. x = or 1 (rejected because x > 0)
1

= .


1

1  2
1 2 1 
1


 
 
14.
Find the last digit of 20142014.
Solution
Since the last digit of a product ab depends only on the last digit of a and of b, then
4=4
4  4 = 16
6  4 = 24
 the last digit repeats with a period of 2.
Since the index 2014 is even, then the last digit of 20142014 is 6.
15.
The diagram shows 9 points. Draw 4 consecutive line segments (i.e. the start point of
the next segment must coincide with the endpoint of the previous segment) to pass
through all the 9 points.
Solution
If you try to draw the line segments within the region bounded by the dots, you will
realise that you need at least 5 consecutive line segments.
 you must draw some of the line segments outside the region as shown:
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Downloaded from studyland.edu.vn
16.
What are the last 5 digits of the sum 1 + 11 + 111 + … + 111…111?
2014 digits
Solution
Last digit: 2014  4
Second last digit: 2013 + 201 = 2214  4
Third last digit: 2012 + 221 = 2233  3
Fourth last digit: 2011 + 223 = 2234  4
Fifth last digit: 2010 + 223 = 2233  3
 the last 5 digits are 34344.
17.
What is the least number of cuts required to cut 10 identical sausages so that they can
be shared equally among 18 people?
Solution
Method 1
Fraction of sausage each person will get = =
This means that there must be at least 10 cuts since no one will get one whole sausage.
Cut each of the 10 sausages at the -mark. Then 10 people will get one sausage each.
There are 10 times sausages left. But each person must get of a sausage.
This means that 2 of the
sausages must be cut into 4 equal parts each (i.e. 3 cuts
each) so that the remaining 8 people will get one sausage and one sausage each.
 least no. of cuts = 10 + 3  2 = 16
Method 2
Least number of cuts required to cut one sausage so that it can be shared equally
among n’ people = n’  1
If m’ and n’ are relatively prime, it can be proven by putting the m’ identical sausages
end to end in one row that the least number of cuts required to cut them so that they
can be shared equally among n’ people is still n’  1, i.e. the cuts will not coincide
with the gaps between the sausages.
If m and n are not relatively prime, some cuts will coincide with the gaps between the
sausages, and this occurs at the end of every set of m’ identical sausages, where m =
m’  HCF(m, n) and n = n’  HCF(m, n), i.e. where m’ and n’ are relatively prime.
Thus the least number of cuts required to cut m identical sausages so that they can
be shared equally among n people is (n’  1)  HCF(m, n) = n  HCF(m, n).
In general, this formula is true for any positive integers m and n, even if m and n are
relatively prime since HCF(m, n) = 1 in the latter case.
 least no. of cuts required to cut 12 identical sausages so that they can be shared
equally among 20 people = 18  HCF(10,18)
= 18  2
= 16
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18.
Divide the following shape into 4 identical parts.
Solution
The shape actually consists of three identical squares.
But 3 and 4 are relatively prime, so it is not easy to divide the three squares into four
identical parts.
So we divide the shape into LCM(3, 4) = 12 parts.
From the 12 parts, we then try to regroup into 4 identical parts as shown below:
19.
Solve the following equation: x5 + 2x3  x2  2 = 0.
Solution
x5 + 2x3  x2  2 = 0
x3(x2 + 2)  (x2 + 2) = 0
(x2 + 2) (x3  1) = 0
Since x2 + 2 > 0, then x3  1 = 0, i.e. x3 = 1.
 x = 1.
Note: If participants solve by guess and check correctly, award 0 mark because they
cannot exclude the possibility that there are other solutions.
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20.
Find the value of
.
Solution
Method 1
1 + 2014 = 2015
2 + 2013 = 2015
3 + 2012 = 2015
There are 1007 pairs of numbers that add up to 2015
1007 + 1008 = 2015
So 1 + 2 + 3 + … + 2014 = 1007  2015

=
=
=
= 1007.5
Method 2
Using the formula 1 + 2 + 3 + … + n =

, 1 + 2 + 3 + … + 2014 =
.
=
=
= 1007.5
21.
In the following cryptarithm, all the letters stand for different digits. Find the values of
A, B, C and D.

A
8
B
C
3
D
9
8
2
0
1
4
Solution
Method 1
In the ones column, C  8 < 4 since C  9, so need to borrow from the tens column.
 10 + C  8 = 4 implies that C = 2.
In the tens column, B  1  9 = B  10 < 1 since B  9, so need to borrow from the
hundreds column.
 10 + B  10 = 1 implies that B = 1.
In the hundreds column, 8  1  D = 0 implies that D = 7.
In the thousands column, A  3 = 2 implies that A = 5.
Check: 5812  3798 = 2014
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Method 2
Rephrase as addition:
+
3
D
9
8
2
0
1
4
A
8
B
C
In the ones column, 8 + 4 = 12, so C = 2.
There is a carryover of 1 from the ones column to the tens column, so 1 + 9 + 1 = 11,
i.e. B = 1.
There is a carryover of 1 from the tens column to the hundreds column, so 1 + D + 0 =
8 or 1 + D + 0 = 18.
If 1 + D + 0 = 18, then D = 17, which is not possible. So 1 + D + 0 = 8, i.e. D = 7.
In the thousands column, 3 + 2 = A, so A = 5.
Check: 5812  3798 = 2014
22.
Find the sum of the terms in the nth pair of brackets:
(1, 2), (3, 4), (5, 6), (7, 8), …
Solution
The sums of the terms in each pair of brackets form the following sequence: 3, 7, 11,
15, …
Method 1
Common difference between consecutive terms, m = 4
Term before the first term, c = 3  4 = 1
 sum of the terms in the nth pair of brackets = mn + c = 4n  1
Method 2
Common difference between consecutive terms, d = 4
First term, a = 3
 sum of the terms in the nth pair of brackets = a + (n  1)d
= 3 + (n  1)  4
= 4n  1
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23.
In the diagram, PQ is parallel to RS, PA = PB and RB = RC. Given that BCA = 60,
find BAC.
A
P
Q
B
60
R
C
S
Solution
Method 1
Let PAB = x. Then PBA = x (base s of isos. ABP)
APB = 180  2x ( sum of ABP)
BRC = 180  APB (corr. s; PQ // RS)
= 180  (180  2x)
= 2x
RBC =
(base s of isos. RBC)
= 90  x
ABC = 180  PBA  RBC (adj. s on a str. ln)
= 180  x  (90  x)
= 90
 BAC = 180  90  60 ( sum of ABC)
= 30
Method 2
Let PAB = x. Then PBA = x (base s of isos. ABP)
APB = 180  2x ( sum of ABP)
BRC = 180  APB (corr. s; PQ // RS)
= 180  (180  2x)
= 2x
RCB =
(base s of isos. RBC)
= 90  x
Now BAC + PAB + APB + BRC + RCB + 60 = 360 ( sum of quad APRC)
BAC + x + (180  2x) + 2x + (90  x) + 60 = 360
 BAC = 30
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24.
Find the remainder when 22014 is divided by 7.
Solution
Method 1
Observe the following pattern: when divided by 7,
21 leaves a remainder of 2,
22 = 4 leaves a remainder of 4,
23 = 8 leaves a remainder of 1,
24 = 16 leaves a remainder of 2,
25 = 32 leaves a remainder of 4,
26 = 64 leaves a remainder of 1, …
This means that the remainder will repeat with a period of 3.
Since 2014 = 671  3 + 1, then 22014, when divided by 7, will leave a remainder of 2.
Method 2
When divided by 7, 21 leaves a remainder of 2.
If 2k leaves a remainder of 2 when divided by 7, it means 2k = 7p + 2 for some integer
p. Then 2k+1 = 2(7p + 2) = 14p + 4 = 7(2p) + 4, i.e. 2k+1 will leave a remainder of 4
when divided by 7.
Let 2k+1 = 7q + 4 for some integer q. Then 2k+2 = 2(7q + 4) = 14q + 8 = 7(2q + 1) + 1,
i.e. 2k+2 will leave a remainder of 1 when divided by 7.
Let 2k+2 = 7r + 1 for some integer r. Then 2k+3 = 2(7r + 1) = 14r + 2 = 7(2r) + 2, i.e.
2k+3 will leave a remainder of 2 when divided by 7.
This means that the remainder will repeat with a period of 3.
Since 21 leaves a remainder of 2 when divided by 7, and 2014 = 671  3 + 1, then
22014, when divided by 7, will leave a remainder of 2.
25.
The diagram shows a triangle ABC where AB = AC, BC = AD and BAC = 20. Find
ADB.
A
20
D
B
C
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Solution
Draw AE perpendicular to BC. Then AE bisects BAC, i.e. BAE = 10 ----- (1)
Draw the point F on AE such that BCF is an equilateral triangle [diagram not drawn
to scale], i.e. FBC = 60.
Draw the line DF.
A
20
D
G
F
B
ABC =
E
C
(base  of isos. ABC) = 80
ABF = ABC  FBC = 80  60 = 20 = BAD (given)
BF = BC (sides of equilateral ) = AD (given)
Since ABF = BAD and BF = AD, then ABFD is an isosceles trapezium [diagram
not drawn to scale].
Let G be the point of intersection of the two diagonals, AF and BD, of the isosceles
trapezium ABFD.
In an isosceles trapezium, AG = BG (by symmetry), i.e. ABG is an isosceles triangle.
ABD = ABG = BAG (base  of isos. ABG) = BAE = 10 from (1)
 ADB = 180  BAD  ABD = 180  20  10 = 150
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SASMO 2014 Round 1 Secondary 2 Problems
1.
The first day of 2014 was a Wednesday. There are 365 days in 2014. In what day of
the week will 2015 begin?
2.
What is the maximum number of parts that can be obtained from cutting a circular
cake using 3 straight cuts?
3.
Evaluate 2014  2014  2013  2015.
4.
Solve
5.
Find the exact value of 1  0.99999…
6.
Mersenne primes are prime numbers of the form Mp = 2p  1, where p is a prime. For
example, 3 = 22  1 is a Mersenne prime. Find the 4th largest Mersenne prime.
7.
Simplify (x  a) (x  b) (x  c) … (x  z).
8.
How many squares are there in a 5  5 square grid?
9.
If x and y are positive integers, find the values of x and y which satisfy the equation
= 3.
x2  4y2 = 41.
10.
Find the dimensions of all the rectangles with integral sides whose area and perimeter
are numerically equal.
11.
A whole number is between 40 and 70. When it is divided by 3, the remainder is 1.
When it is divided by 7, the remainder is 2. Find the number.
12.
Find the value of
2

2
2  12
 2  12 2
2

 




.
13.
Find the value of
.
14.
Find the sum of the terms in the nth pair of brackets:
(1, 2, 3, 4), (5, 6, 7, 8), (9, 10, 11, 12), …
15.
In the diagram, PQ and RS are parallel, PA = PB and RA = RC. Find BAC.
P
B
A
R
C
S
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16.
There are 367 students in the school hall. What is the probability that two of the
students have their birthday falling on the same day of the year?
17.
Let x be a number such that x 
18.
Divide the following shape into 4 identical parts.
19.
Find all the solutions to the trigonometric equation sin2 x  cos2 x = 1 for 0  x  360.
20.
The diagram shows a square AEFG with an inscribed circle. ABCD is a rectangle such
that AB = 2 cm and AD = 4 cm. Find the radius of the circle.
1
1
 5 . Find the value of x 3  3 .
x
x
A
B
D
C
G
21.
E
F
In the following cryptarithm, all the letters stand for different digits. Find the final
four-digit product.

+
*
H
E
E
H
*
8
*
*
*
*
*
22.
A farmer wants to plant 10 trees in 5 rows such that there are exactly 4 trees in each
row. Draw a diagram to show how the trees should be planted.
23.
Find the last three digits of 2525.
24.
Find the remainder when 32014 is divided by 5.
25.
Find the largest number of composite numbers less than 2014 that are relatively prime
to one another, i.e. the highest common factor of any two of them is 1.
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Downloaded from studyland.edu.vn
SASMO 2014 Round 1 Secondary 2 Solutions
1.
The first day of 2014 was a Wednesday. There are 365 days in 2014. In what day of
the week will 2015 begin?
Solution
365 days = 52 weeks and 1 day
Since the first day of 2014 was a Wednesday, then the first day of 2015 will be
Thursday.
2.
What is the maximum number of parts that can be obtained from cutting a circular
cake using 3 straight cuts?
Solution
Maximum number of parts = 8
First Cut
Second Cut
Third Cut
3.
Evaluate 2014  2014  2013  2015.
Solution
Method 1
2014  2014  2013  2015
= (2013 + 1)  2014  2013  (2014 + 1)
= 2013  2014 + 2014  2013  2014  2013
= 2014  2013
=1
Method 2
2014  2014  2013  2015
= 2014  2014  (2014  1)  (2014 + 1)
= 2014  2014  (2014  2014  1)
=1
4.
Solve
= 3.
Solution
=3
=9
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Downloaded from studyland.edu.vn
x+3=9
x=6
5.
Find the exact value of 1  0.99999…
Solution
Let x = 0.99999…
Then 10x = 9.99999…
 9x = 10x – x = 9
x= =1
 0.99999… = 1
i.e. 1  0.99999… = 0
Note: 0.00000… is not acceptable since it suggests that the participant does not
know that the answer is exactly 0.
6.
Mersenne primes are prime numbers of the form Mp = 2p  1, where p is a prime. For
example, 3 = 22  1 is a Mersenne prime. Find the 4th largest Mersenne prime.
Solution
M2 = 22  1 = 3 is the smallest Mersenne prime.
M3 = 23  1 = 7 is prime.
M5 = 25  1 = 31 is prime.
M7 = 27  1 = 127 is prime.
 the 4th largest Mersenne prime is 127.
7.
Simplify (x  a) (x  b) (x  c) … (x  z).
Solution
Since x  x = 0, then (x  a) (x  b) (x  c) … (x  z) = 0.
8.
How many squares are there in a 5  5 square grid?
Solution
No. of 1  1 squares = 25 = 52
No. of 2  2 squares = 16 = 42
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No. of 3  3 squares = 9 = 32
No. of 4  4 squares = 4 = 22
No. of 5  5 squares = 1 = 12
 total no. of squares in a 5  5 square grid = 12 + 22 + 32 + 42 + 52 = 55
9.
If x and y are positive integers, find the values of x and y which satisfy the equation
x2  4y2 = 41.
Solution
x2  4y2 = 41
(x + 2y)(x  2y) = 41
Since 41 is prime, there are only two factors of 41, i.e. 1 and 41.
Since x and y are positive integers and x + 2y > x  2y, then x + 2y = 41 and x  2y = 1.
Solving, x = 21 and y = 10.
10.
Find the dimensions of all the rectangles with integral sides whose area and perimeter
are numerically equal.
Solution
Method 1
Let the dimensions of a rectangle be l units by b units.
Then area = perimeter
lb  2l  b
2b
l
b2
Since b  2 > 0, then b > 2.
When b = 3, l = 6.
When b = 4, l = 4.
When b = 5, l = is not an integer.
When b = 6, l = 3, which is essentially the same as b = 3, l = 6.
If b > 6, l < 3.
2l
By symmetry, b 
, i.e. l > 2.
l2
So there are no other solutions.
 the dimensions of all the rectangles with integral sides whose area and perimeter are
numerically equal are 3 by 6 and 4 by 4.
Method 2
Let the dimensions of a rectangle be l units by b units.
Then area = perimeter
lb  2l  b
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2b
b2
2b  4  4

b2
2b  2  4

b2
4
 2
b2
Since b  2 > 0, then b > 2.
4
For l to be an integer,
must also be an integer.
b2
l
This means that 4  b – 2, i.e. b  6.
So the only possible solutions are when b = 3, 4, 5 and 6.
When b = 3, l = 6.
When b = 4, l = 4.
When b = 5, l = is not an integer.
When b = 6, l = 3, which is essentially the same as b = 3, l = 6.
 the dimensions of all the rectangles with integral sides whose area and perimeter are
numerically equal are 3 by 6 and 4 by 4.
11.
A whole number is between 40 and 70. When it is divided by 3, the remainder is 1.
When it is divided by 7, the remainder is 2. Find the number.
Solution
Method 1
Since the number leaves a remainder of 1 when divided by 3, then the possible values
of the number are 40, 43, 46, 49, 52, 55, 58, …, 70.
Since the number leaves a remainder of 2 when divided by 7, we can either check each
of the above possible values one by one, or we can start with 7  6 + 2 = 44, and then
add 7 until we get a number in the first list, i.e. 44, 51, 58, …
 the number is 58.
Method 2
Let the number be N. Then N  1 (mod 3) and N  2 (mod 7).
Since the moduli 3 and 7 are relatively prime, we can use the Chinese Remainder
Theorem as followed.
7N  1 (mod 3)  N  1 (mod 3)  N 1 = 4;
3N  2 (mod 7)  N 2 = 3.
Then N  7  4 + 3  3 = 37 (mod 21).
Since N is between 40 and 70, then N = 37 + 21 = 58.
12.
Find the value of
2

2
2  12
 2  12 2
2

 




.
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Solution
2
Let
= x. Then
2
= x, i.e. 12x2 + 2x  2 = 0.
2  12 x


2

2  12
 2  12 2 
2


2
 6x + x  1 = 0 implies (3x  1) (2x + 1) = 0, i.e. x = or  (rejected because x > 0)
2

= .


2

2  12
 2  12 2 
2


 
 
13.
Find the value of
.
Solution
Method 1
1 + 2013 = 2014
3 + 2011 = 2014
5 + 2009 = 2014
There are
= 503 pairs of numbers that add up to 2014
1005 + 1009 = 2014
1007 = 1007
So 1 + 3 + 5 + … + 2013 = 503  2014 + 1007

=
=
= 503 +
= 503.5
Method 2
1 + 2013 = 2014
3 + 2011 = 2014
5 + 2009 = 2014
2013 + 1
There are
= 1007 pairs of numbers that add up to 2014
= 2014
= 1007  1007
So 1 + 3 + 5 + … + 2013 =

=
=
=
= 503.5
14.
Find the sum of the terms in the nth pair of brackets:
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(1, 2, 3, 4), (5, 6, 7, 8), (9, 10, 11, 12), …
Solution
The sums of the terms in each pair of brackets form the following sequence: 3, 7, 11,
15, …
Method 1
Common difference between consecutive terms, m = 4
Term before the first term, c = 3  4 = 1
 sum of the terms in the nth pair of brackets = mn + c = 4n  1
Method 2
Common difference between consecutive terms, d = 4
First term, a = 3
 sum of the terms in the nth pair of brackets = a + (n  1)d
= 3 + (n  1)  4
= 4n  1
15.
In the diagram, PQ and RS are parallel, PA = PB and RA = RC. Find BAC.
P
B
A
R
C
S
Q
Solution
Let PAB = x. Then PBA = x (base s of isos. ABP)
APB = 180  2x ( sum of ABP)
ARC = 180  APB (corr. s; PQ // RS)
= 180  (180  2x)
= 2x
RAC =
(base s of isos. ARC)
= 90  x
 BAC = 180  PAB  RAC (adj. s on a str. ln)
= 180  x  (90  x)
= 90
16.
There are 367 students in the school hall. What is the probability that two of the
students have their birthday falling on the same day of the year?
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Solution
There are at most 366 days in a year.
Suppose none of the first 366 students have their birthday on the same day of the year.
Then the last student’s birthday must fall on the same day of the year as the birthday
of one of the first 366 students.
 probability two of the students have their birthday on the same day of the year = 1
17.
Let x be a number such that x 
1
1
 5 . Find the value of x 3  3 .
x
x
Solution
2
1

x 
x

 2 1
x  2
x

18.
1
1
= 25  x 2  2 = 23
2
x
x
1 1
1
1
1 
1

 2 1 
3
3
 x   = x  x   3  x  3 =  x  2  x     x  
x x
x
x 
x
x
x 


1 
1


=  x   x 2  2  1
x 
x


= 5  (23  1)
= 110
= 52  x 2  2 
Divide the following shape into 4 identical parts.
Solution
The shape actually consists of three identical squares.
But 3 and 4 are relatively prime, so it is not easy to divide the three squares into four
identical parts.
So we divide the shape into LCM(3, 4) = 12 parts first.
From the 12 parts, we then try to regroup into 4 identical parts as shown below:
19.
Find all the solutions to the trigonometric equation sin2 x  cos2 x = 1 for 0  x  360.
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Solution
Since sin2 x  1 and cos2 x  0, then the only possible solution to sin2 x  cos2 x = 1 is
when sin2 x = 1 and cos2 x = 0, i.e. when sin x = 1 and cos x = 0.
 x = 90 or 270
20.
The diagram shows a square AEFG with an inscribed circle. ABCD is a rectangle such
that AB = 2 cm and AD = 4 cm. Find the radius of the circle.
A
B
D
C
E
G
F
Solution
Draw CJ and OH parallel to AE, and OI parallel to AG. Then OCJ is a right-angled
triangle.
I
A B
E
D
H
G
C
J
O
F
Let the radius of the circle be r cm. Then OC = OH = OI = r.
So CJ = r  2 and OJ = r  4.
By Pythagoras’ Theorem, OC 2 = CJ 2 + OJ 2, i.e. r2 = (r  2)2 + (r  4)2.
Simplifying, we have r2  12r + 20 = 0, i.e. (r  2) (r  10) = 0.
Since r > 4, then the radius of the circle is 10 cm.
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21.
In the following cryptarithm, all the letters stand for different digits. Find the final
four-digit product.

+
*
H
E
E
H
*
8
*
*
*
*
*
Solution
Replace the * with letters, which can be the same.

+
D
H
E
E
H
A
8
B
C
F
G
I
First I = 8.
Since the addition of two digits will give a maximum of 18, or a maximum of 19 if
there is a carryover (or renaming) of 1, this means that the maximum carryover is 1.
So B = 9 and there is a carryover of 1 for B to give 10, i.e. D = 1 and F = 0.
For HE  H = A8 to be a two-digit number, H < 4.
If H = 1, E = 8; if H = 2, E = 4 or 9; if H = 3, E = 6.
For HE  E = BC to be a two-digit number,
if H = 1, then E  6, so E  8;
if H = 3, then E  2, so E  6;
if H = 2, then E  4, so E  9, but E = 4.
So H = 2 and E = 4.
 the final four-digit product is HE  EH = 24  42 = 1008
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22.
A farmer wants to plant 10 trees in 5 rows such that there are exactly 4 trees in each
row. Draw a diagram to show how the trees should be planted.
Solution
If you start with 10 trees and then try fitting 5 lines onto them, you will realise that the
lines must overlap because there are not enough trees, and it’s not easy to fit 5 lines
onto 10 trees.
 try to draw 5 overlapping lines first, and a common figure with 5 overlapping lines
is the following star:
Then put in the trees and yes, it works.
23.
Find the last three digits of 2525.
Solution
Since the last three digits of a product ab depends only on the last three digits of a and
of b, then
25 = 025
025  25 = 625
625  25 = 15 625
 the last three digits are always 625, with the exception of the first power.
 the last three digits of 2525 are 625.
24.
Find the remainder when 32014 is divided by 5.
Solution
Method 1
Observe the following pattern: when divided by 5,
31 leaves a remainder of 3,
32 = 9 leaves a remainder of 4,
33 = 27 leaves a remainder of 2,
34 = 81 leaves a remainder of 1,
35 = 243 leaves a remainder of 3,
36 = 729 leaves a remainder of 4,
37 = 2187 leaves a remainder of 2,
38 = 6561 leaves a remainder of 1, …
This means that the remainder will repeat with a period of 4.
Since 2014 = 503  4 + 2, then 32014, when divided by 5, will leave a remainder of 4.
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Method 2
When divided by 5, 31 leaves a remainder of 3.
If 3k leaves a remainder of 3 when divided by 5, it means 3k = 5p + 3 for some integer
p. Then 3k+1 = 3(5p + 3) = 15p + 9 = 15p + 5 + 4 = 5(3p + 1) + 4, i.e. 3k+1 will leave
a remainder of 4 when divided by 5.
Let 3k+1 = 5q + 4 for some integer q. Then 3k+2 = 3(5q + 4) = 15q + 12 = 15q + 10 + 2
= 5(3q + 2) + 2, i.e. 3k+2 will leave a remainder of 2 when divided by 5.
Let 3k+2 = 5r + 2 for some integer r. Then 3k+3 = 3(5r + 2) = 15r + 6 = 15r + 5 + 1 =
5(3r + 1) + 1, i.e. 3k+3 will leave a remainder of 1 when divided by 5.
Let 3k+3 = 5s + 1 for some integer s. Then 3k+4 = 3(5s + 1) = 15s + 3, i.e. 3k+4 will leave
a remainder of 3 when divided by 5.
This means that the remainder will repeat with a period of 4.
Since 31 leaves a remainder of 3 when divided by 4, and 2014 = 503  4 + 2, then
22014, when divided by 4, will leave a remainder of 4.
25.
Find the largest number of composite numbers less than 2014 that are relatively prime
to one another, i.e. the highest common factor of any two of them is 1.
Solution
If 6 = 2  3 is one of the composite numbers, then all the multiples of 2 and all the
multiples of 3 will not be relatively prime to 6.
Thus, to get the largest number of composite numbers, we have to choose 2n and 3m
instead of 6k.
In other words, we have to choose pn, where p is prime.
Since the composite numbers must be less than 2014, then n must not be too big, esp.
when p is big enough.
Now, the largest value for p is 43, since 432 = 1849 < 2014 but 472 = 2209 > 2014.
Without loss of generality, to get the largest number of composite numbers less than
2014 that are relatively prime to one another, choose the following numbers: 22, 32,
52, 72, …, 432 (although we can choose, e.g. 210 = 1024 < 2014 instead of 22).
 largest number of composite numbers less than 2014 that are relatively prime to one
another = number of primes  43
= 14
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