Division
S2
Singapore and Asian
Schools Math Olympiad
2015
Full Name:
Index Number:
Class:
School:
SASMO 2015 Secondary 2 Contest
INSTRUCTIONS
1. Please DO NOT OPEN the contest booklet until the Proctor has given permission to
start.
2. TIME: 1 hour 30 minutes.
3. Attempt all 25 questions.
Questions 1 to 15 score 2 points each, no points are deducted for unanswered
question and 1 point is deducted for wrong answer.
Questions 16 to 25 score 4 points each. No points are deducted for unanswered or
wrong answers.
4. Shade your answers neatly using a pencil in the answer sheet.
5. PROCTORING: No one may help any student in any way during the contest.
6. No electronic devices capable of storing and displaying visual information is
allowed during the course of the exam.
7. Strictly No Calculators are allowed into the exam.
8. All students must fill and shade in their Name, Index number, Class and School in
the answer sheet and contest booklet.
9. MINIMUM TIME: Students must stay in the exam hall at least 1h 15 min.
10. Students must show detailed working and transfer answers to the answer sheet.
11. No exam papers and written notes can be taken out by any contestant.
10
SASMO 2015, Secondary 2 Contest
SASMO 2015 Secondary 2
Starting Score = 15 marks (to avoid negative marks); Max Possible Score = 85
marks
Section A (Correct answer = 2 marks; no answer = 0; incorrect answer =
minus 1 mark)
1.
The diagram shows a big rectangle that is made up of 4 small identical
rectangles. If the smaller dimension of one of the small rectangles is 9 cm,
find the area of the big rectangle.
(a)
(b)
(c)
(d)
(e)
108 cm2
972 cm2
2015 cm2
Not enough given information to find
None of the above
________________________________________________________________
2.
Find the next term of the following sequence: 3, 2, 5, 9, 44, …
(a)
(b)
(c)
(d)
(e)
86
87
88
395
2015
12
SASMO 2015, Secondary 2 Contest
3.
An operator  acts on two numbers to give the following outcomes:
3
5
1
4




2
3
6
9
=
94
= 259
= 136
= 1681
What is 5  7 equal to?
(a)
(b)
(c)
(d)
(e)
2549
3549
4925
4935
None of the above
________________________________________________________________
4.
If the five-digit number 4567N is divisible by 12, find N.
(a)
(b)
(c)
(d)
(e)
8
6
4
2
0
13
SASMO 2015, Secondary 2 Contest
5.
The diagram shows a square, a regular hexagon and parts of a regular
polygon. How many sides does the latter polygon have?
(a)
(b)
(c)
(d)
(e)
10
12
20
24
30
________________________________________________________________
6.
A palindromic number is a whole number that reads the same forward and
backward. For example, 1221 is a palindromic number. How many 6-digit
palindromic numbers are there?
(a)
(b)
(c)
(d)
(e)
190
900
1000
90000
None of the above
14
SASMO 2015, Secondary 2 Contest
7.
A farmer buys 36 metres of fence to build an enclosure to keep his livestock.
He decides to erect a 4-sided enclosure in such a way that it has the largest
possible area. Find the area of the enclosure.
(a)
(b)
(c)
(d)
(e)
63 m2
72 m2
81 m2
90 m2
None of the above
________________________________________________________________
8.
A number gives a remainder of 5 when divided by 10. Another number gives
a remainder of 4 when divided by 10. The sum of these two numbers is
multiplied by 3 to give the third number. What is the remainder when this
third number is divided by 10?
(a)
(b)
(c)
(d)
(e)
4
5
7
9
None of the above
15
SASMO 2015, Secondary 2 Contest
9.
A big cube is made up of 64 small cubes. All the faces of the big cube are
then painted. How many of the small cubes have exactly one painted face?
(a)
(b)
(c)
(d)
(e)
6
12
24
72
None of the above
________________________________________________________________
10.
If both n and
there?
(a)
(b)
(c)
(d)
(e)
108
𝑛3
are positive integers, how many possible values of n are
1
2
3
4
None of the above
16
SASMO 2015, Secondary 2 Contest
11.
The product of two numbers is 100 000. Neither of the two numbers has 10
as a factor. Find the difference of these two numbers.
(a)
(b)
(c)
(d)
(e)
465
1170
3093
6234
Not enough given information to find
________________________________________________________________
12.
The height of a man is 169 cm, correct to the nearest centimetre. What is the
upper bound of the man’s height?
(a)
(b)
(c)
(d)
(e)
169.4 cm
169.45 cm
169.49 cm
169.5 cm
169.9 cm
17
SASMO 2015, Secondary 2 Contest
13.
A circle and a rhombus are drawn on a flat surface. What is the biggest
number of regions that can be formed on the surface?
(a)
(b)
(c)
(d)
(e)
6
7
8
9
10
________________________________________________________________
14.
Find the smallest whole number n for which 540n is a multiple of 168.
(a)
(b)
(c)
(d)
(e)
2
7
14
28
None of the above
18
SASMO 2015, Secondary 2 Contest
15.
In ABC, AB = 10 cm, BC = 6 cm and AC = 8 cm. Find the value of
(a)
(b)
(c)
(d)
(e)
sin 𝐴
sin 𝐵
.
0.6
0.75
0.8
Cannot be found
None of the above
________________________________________________________________
Section B (Correct answer = 4 marks; incorrect or no answer = 0)
16.
There are 15 rows of seats in a hall, each row having 20 seats. If there are
275 people seated in the hall, at least how many rows have an equal number
of people each?
19
SASMO 2015, Secondary 2 Contest
17.
Tom drives 4000 km during a trip. He rotates the tyres (four tyres on the car
and one spare tyre) so that each tyre has been used for the same distance at
the end of the trip. How many kilometres are covered by each tyre?
________________________________________________________________
18.
The diagram shows how an equilateral triangle can be cut into four pieces
and rearranged to form a square. This solution of the Haberdasher’s Puzzle is
discovered by Henry Dudeney (1857 – 1930).
If the length of the square is 13 cm, find the length of the triangle, correct to
2
nearest whole number. (Note: 4 = 1.52, correct to three significant figures)
√3
20
SASMO 2015, Secondary 2 Contest
19.
Polite numbers are numbers that can be expressed as the sum of two or more
consecutive positive integers, e.g.
5 = 2 + 3;
9 = 2 + 3 + 4 = 4 + 5.
Find all the polite numbers that can be expressed as the sum of three
consecutive positive integers.
________________________________________________________________
20.
Find the 2015th term of the following sequence: 1, 2, 2, 3, 3, 3, 4, 4, 4, 4, …
21
SASMO 2015, Secondary 2 Contest
21.
The diagram shows a rectangle being divided into 3 smaller rectangles and a
square. If the perimeter of the unshaded rectangle is 54 cm and the area of
the square is 64 cm2, find the total area of the shaded rectangles.
________________________________________________________________
22.
Find the value of √2 + √2 + √2+. . .
22
SASMO 2015, Secondary 2 Contest
23.
How many rectangles are there in a 6  6 square grid?
________________________________________________________________
24.
In the following cryptarithm, all the different letters stand for different digits.
Find the 5-digit number EFCBH.
+
E
A
B
C
D
E
F
G
B
F
C
B
H
23
SASMO 2015, Secondary 2 Contest
25.
Find the last five digits of 152015.
END OF PAPER
24
SASMO 2015, Secondary 2 Contest
SASMO 2015 Secondary 2 Solution
[20 MCQ + 5 = 25 Q]
Section A
1.
The diagram shows a big rectangle that is made up of 4 small identical
rectangles. If the smaller dimension of one of the small rectangles is 9 cm,
find the area of the big rectangle.
(a)
(b)
(c)
(d)
(e)
108 cm2
972 cm2 [Ans]
2015 cm2
Not enough given information to find
None of the above
Solution
Since the smaller dimension of the small rectangle is 9 cm, then the bigger
dimension of the small rectangle is 27 cm as shown:
9
9
9
27
So the length of the big rectangle is 27 + 9 = 36 cm
 area of big rectangle = 36  27 = 972 cm2
2.
Find the next term of the following sequence: 3, 2, 5, 9, 44, …
(a)
(b)
(c)
(d)
(e)
86
87
88
395 [Ans: 9  44  1]
2015
Solution
From the third term onwards, the next term is obtained by multiplying the
previous two terms and then subtracting 1.
 the next term is 9  44  1 = 395.
25
SASMO 2015, Secondary 2 Contest
3.
An operator  acts on two numbers to give the following outcomes:
3
5
1
4




2
3
6
9
=
94
= 259
= 136
= 1681
What is 5  7 equal to?
(a)
(b)
(c)
(d)
(e)
2549 [Ans]
3549
4925
4935
None of the above
Solution
a  b = (𝑎 × 𝑎)(𝑏 × 𝑏)
 5  7 = 3549
4.
If the five-digit number 4567N is divisible by 12, find N.
(a)
(b)
(c)
(d)
(e)
8
6
4
2 [Ans]
0
Solution
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 3, 4 + 5 + 6 + 7 + N = N + 22 is also divisible
by 3.
Since N is a single digit, N = 2, 5 and 8.
Using the divisibility test for 4, the last two digits 7N is also divisible by 4.
Since 75 and 78 are not divisible by 4, then N = 2.
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SASMO 2015, Secondary 2 Contest
5.
The diagram shows a square, a regular hexagon and parts of a regular
polygon. How many sides does the latter polygon have?
(a)
(b)
(c)
(d)
(e)
10
12 [Ans]
20
24
30
Solution
Method 1
Sum of interior angles of regular hexagon = (n  2)  180
= (6  2)  180
= 4  180
= 720
Each interior angle of regular hexagon = 720  6
= 120
Each interior angle of square = 90
Each interior angle of regular polygon = 360  120  90
= 150
Let the no. of sides of the regular polygon be m.
Then (m  2) × 180 = 150 × m
180m  360 = 150m
30m = 360
m = 12
 the regular polygon has 12 sides.
Method 2
Each exterior angle of regular hexagon = 360  6 (since sum of ext. angles
= 360)
= 60
Each interior angle of regular hexagon = 180  60
= 120
Each interior angle of square = 90
Each interior angle of regular polygon = 360  120  90
= 150
Each exterior angle of regular polygon = 180  150
= 30
No. of sides of regular polygon = no. of exterior angles of regular polygon
= 360  30
= 12
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SASMO 2015, Secondary 2 Contest
6.
A palindromic number is a whole number that reads the same forward and
backward. For example, 1221 is a palindromic number. How many 6-digit
palindromic numbers are there?
(a)
(b)
(c)
(d)
(e)
190
900 [Ans]
1000
90000
None of the above
Solution
There are only 9 possibilities for the hundred thousands digit because the
hundred thousands digit of a 5-digit number cannot be 0.
Since the ones digit is the same as the hundred thousands digit in a 6-digit
palindromic number, there are no other possibilities for the ones digit once
the hundred thousands digit is fixed.
However, there are 10 possibilities for the ten thousands digit.
Since the tens digit is the same as the ten thousands digit in a 6-digit
palindromic number, there are no other possibilities for the tens digit once
the ten thousands digit is fixed.
Similarly, there are 10 possibilities for the thousands digit.
Since the hundreds digit is the same as the thousands digit in a 6-digit
palindromic number, there are no other possibilities for the hundreds digit
once the thousands digit is fixed.
 there are 9 × 10 × 10 = 900 six-digit palindromic numbers.
7.
A farmer buys 36 metres of fence to build an enclosure to keep his livestock.
He decides to erect a 4-sided enclosure in such a way that it has the largest
possible area. Find the area of the enclosure.
(a)
(b)
(c)
(d)
(e)
63 m2
72 m2
81 m2 [Ans]
90 m2
None of the above
Solution
A 4-sided enclosure has the largest possible area when it is a square.
Length of enclosure = 36 × 4 = 9 m
 area of enclosure = 9 m × 9 m = 81 m2
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SASMO 2015, Secondary 2 Contest
8.
A number gives a remainder of 5 when divided by 10. Another number gives
a remainder of 4 when divided by 10. The sum of these two numbers is
multiplied by 3 to give the third number. What is the remainder when this
third number is divided by 10?
(a)
(b)
(c)
(d)
(e)
4
5
7 [Ans: (5 + 4)  3 = 27]
9
None of the above
Solution
Method 1
1st number = 10 + 10 + 10 + … + 10 + 5 (it does not matter how many 10s
are there)
2nd number = 10 + 10 + 10 + … + 10 + 4 (it does not matter how many 10s
are there)
Sum of first two numbers = 10 + 10 + 10 + … + 10 + 9
3rd number = 3 × sum of first two numbers
= 10 + 10 + 10 + … + 10 + 27 (it does not matter how many 10s
are there)
= 10 + 10 + 10 + … + 10 + 10 + 10 + 7
 when the 3rd number is divided by 10, the remainder is 7.
Method 2
Let the first number be 10x + 5 and the second number be 10y + 4, where x
and y are whole numbers.
Then the sum of the first two numbers is (10x + 5) + (10y + 4) = 10(x + y)
+9
= 10z + 9 for some whole number z.
Thus the third number is 3(10z + 9) = 30z + 27 = 30z + 20 + 7 = 10(3z + 2)
+7
= 10w + 7 for some whole number w.
 when the third number is divided by 10, the remainder is 7.
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SASMO 2015, Secondary 2 Contest
9.
A big cube is made up of 64 small cubes. All the faces of the big cube are
then painted. How many of the small cubes have exactly one painted face?
(a)
(b)
(c)
(d)
(e)
6
12
24 [Ans: 4 × 6 faces = 24]
72
None of the above
Solution
Since 64 = 4 × 4 × 4, then the dimensions of the big cube are 4 by 4 by 4.
The only small cubes that have exactly one painted face are those that are
not along the edges of the big cube.
There are 4 such small cubes on each face of the big cube.
Since the big cube had 6 faces, then there are 4 × 6 = 24 small cubes have
exactly one painted face.
10.
108
If both n and
𝑛3
there?
(a)
(b)
(c)
(d)
(e)
are positive integers, how many possible values of n are
1
2 [Ans: n = 1 or 3]
3
4
None of the above
Solution
Method 1
If both n and
108
𝑛3
are positive integers, then n3 is a factor of 108.
108 = 1 × 108
= 2 × 54
= 3 × 36
= 4 × 27
= 6 × 18
= 9 × 12
Since n3 is a perfect cube, the possible values of n3 are 1 and 27, i.e. n = 1 or
3.
 there are 2 possible values of n.
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SASMO 2015, Secondary 2 Contest
Method 2
If both n and
108
𝑛3
are positive integers, then n3 is a factor of 108.
108 = 2 × 2 × 3 × 3 × 3
Since n3 is a perfect cube, the possible values of n3 are 13 and 33, i.e. n = 1
or 3.
 there are 2 possible values of n.
11.
The product of two numbers is 100 000. Neither of the two numbers has 10
as a factor. Find the difference of these two numbers.
(a)
(b)
(c)
(d)
(e)
465
1170
3093 [Ans: 3125  32 = 3093]
6234
Not enough given information to find
Solution
100 000 = 2 × 2 × 2 × 2 × 2 × 5 × 5 × 5 × 5 × 5
Since 2 × 5 = 10, each of the two numbers must not contain both a 2 and a 5
as its factors, or else it will contain 10 as its factor.
Thus 100 = (2 × 2 × 2 × 2 × 2) × (5 × 5 × 5 × 5 × 5) = 32 × 3125.
 the two numbers are 32 and 3125, and their difference is 3125  32 =
3093.
12.
The height of a man is 169 cm, correct to the nearest centimetre. What is the
upper bound of the man’s height?
(a)
(b)
(c)
(d)
(e)
169.4 cm
169.45 cm
169.49 cm
169.5 cm [Ans]
169.9 cm
Solution
Let the man’s height be h cm.
Since 169 cm is correct to the nearest centimetre, then 168.5  h < 169.5.
 the upper bound of the man’s height is 169.5 cm.
Note: It is not necessary for the upper bound to be in the interval 168.5  h
< 169.5.
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SASMO 2015, Secondary 2 Contest
13.
A circle and a rhombus are drawn on a flat surface. What is the biggest
number of regions that can be formed on the surface?
(a)
(b)
(c)
(d)
(e)
6
7
8
9
10 [Ans]
Solution
To form the biggest number of regions, the rhombus and the circle should
intersect as often as possible.
Since a line can cut a circle at most two times, the question is whether it is
possible for each of the four line segments of a rhombus to cut the circle
two times.
The following diagram shows that it is possible and so the biggest number of
regions that can be formed on the surface is 10.
4
3
2
5
9
1
8
6
10
7
Note: Since a square is also a rhombus, the above diagram can be drawn
using a square too.
14.
Find the smallest whole number n for which 540n is a multiple of 168.
(a)
(b)
(c)
(d)
(e)
2
7
14 [Ans]
28
None of the above
32
SASMO 2015, Secondary 2 Contest
Solution
Method 1
Since 540n is a common multiple of 540 and 168, and n is the smallest whole
number, then 540n is the LCM of 540 and 168.
540 = 22 × 33 × 5
168 = 23 × 3 × 7
 LCM of 540 and 168 = 23 × 33 × 5 × 7
Since 540n is the LCM of 540 and 168, then 540n = 23 × 33 × 5 × 7
22 × 33 × 5 × n = 23 × 33 × 5 × 7
n=2×7
= 14
Method 2
Since 540n is a common multiple of 540 and 168, and n is the smallest whole
number, then 540n is the LCM of 540 and 168.
540n = 22 × 33 × 5 × n
168 = 23 × 3 × 7
Since 540n is the LCM of 540 and 168, then n = 2 × 7 = 14.
Method 3
540n is a multiple of 168 means 540n is divisible by 168, i.e.
number.
In other words, 540n must contain all the prime factors of 168.
Since
15.
540𝑛
168
=
540𝑛
168
22 ×33 ×5×𝑛
23 ×3×7
, , then the smallest whole number n is 2 × 7 = 14.
In ABC, AB = 10 cm, BC = 6 cm and AC = 8 cm. Find the value of
(a)
(b)
(c)
(d)
(e)
is a whole
sin 𝐴
sin 𝐵
.
0.6
0.75 [Ans]
0.8
Cannot be found
None of the above
Solution
Since 62 + 82 = 102, ABC is a right-angled triangle with C = 90 (converse
of Pythagoras’ Theorem).
opp
BC
6
3
opp
AC
8
4
Then sin A = hyp = AB = 10 = 5 and sin B = hyp = AB = 10 = 5.
33
SASMO 2015, Secondary 2 Contest

sin 𝐴
sin 𝐵
=
3
5

4
5
=
3
5

5
4
=
𝟑
𝟒
or 0.75
Section B
16.
There are 15 rows of seats in a hall, each row having 20 seats. If there are
275 people seated in the hall, at least how many rows have an equal number
of people each?
Solution
Worst case scenario: try to squeeze as many people into the seats with as
few rows as possible with equal number of people.
Suppose no rows have an equal number of people.
Then there are 20 + 19 + 18 + … + 6 = 195 people, but there are 275
people.
Suppose at least 2 rows have an equal number of people.
Then there are (20 + 19 + 18 + … + 14)  2 + 13 = 251 people, but there
are 275 people.
Suppose at least 3 rows have an equal number of people.
Then there are (20 + 19 + 18 + 17 + 16)  3 = 270 people, but there are
275 people.
Suppose at least 4 rows have an equal number of people.
Then there are (20 + 19 + 18)  4 + 17 + 16 + 15 = 276 people, which is
more than 275 people.
 there are at least 4 rows with equal number of people.
17.
Tom drives 4000 km during a trip. He rotates the tyres (four tyres on the car
and one spare tyre) so that each tyre has been used for the same distance at
the end of the trip. How many kilometres are covered by each tyre?
Solution
Since there are 4 tyres on the road at any one time, total distance covered by
4 tyres = 4000 km  4 = 16 000 km
Since 5 tyres share the total distance equally, then distance covered by each
tyre = 16 000 km  5 = 3200 km
34
SASMO 2015, Secondary 2 Contest
18.
The diagram shows how an equilateral triangle can be cut into four pieces
and rearranged to form a square. This solution of the Haberdasher’s Puzzle is
discovered by Henry Dudeney (1857 – 1930).
If the length of the square is 13 cm, find the length of the triangle, correct to
2
nearest whole number. (Note: 4 = 1.52, correct to three significant figures)
√3
Solution
Let the length of the square be l cm, the length of the triangle be b cm and
the height of the triangle be h cm.
1
Since area of triangle = area of square, then 2bh = l 2, i.e. h =
𝑏 2
Using Pythagoras’ Theorem, b 2 = h 2 + (2) , i.e. b 2 =
4 2𝑙2
Subst. (1) into (2), b 2 = 3 (
4
16
 b = √3 l =
2
4
√3
𝑏
2
4 4𝑙4
) = 3 ( 𝑏2 ), i.e. b 4 =
16
3
4
3
2𝑙2
𝑏
--- (1)
h 2 --- (2)
l 4.
l = 1.52  13 = 20 cm (to nearest whole number).
35
SASMO 2015, Secondary 2 Contest
19.
Polite numbers are numbers that can be expressed as the sum of two or more
consecutive positive integers, e.g.
5 = 2 + 3;
9 = 2 + 3 + 4 = 4 + 5.
Find all the polite numbers that can be expressed as the sum of three
consecutive positive integers.
Solution
Method 1
1+2+3=6
2+3+4=9
3 + 4 + 5 = 12
⋮
Each polite number is obtained from the previous one by adding 3.
 all the polite numbers that can be expressed as the sum of three
consecutive positive integers are all the multiples of 3 except 3.
Note: It is understood that multiples of 3 are positive unless otherwise stated.
Method 2
All the polite numbers that can be expressed as the sum of three consecutive
positive integers are of the form n + (n + 1) + (n + 2) = 3n + 3, where n
is a positive integer.
 all the polite numbers that can be expressed as the sum of three
consecutive positive integers are all the multiples of 3 except 3.
20.
Find the 2015th term of the following sequence: 1, 2, 2, 3, 3, 3, 4, 4, 4, 4, …
Solution
Notice that there are n terms with values equal to n each.
So find the largest value of n such that 1 + 2 + 3 + … + n  2015 first.
𝑛(𝑛+1)
 2  2015, i.e. n(n + 1)  4030.
Since 62  63 = 3906 and 63  64 = 4032, then the largest value of n is 62.
Since 1 + 2 + 3 + … + 62 < 2015 and 1 + 2 + 3 + … + 63 > 2015, then the
2015th term must be 63.
36
SASMO 2015, Secondary 2 Contest
21.
The diagram shows a rectangle being divided into 3 smaller rectangles and a
square. If the perimeter of the unshaded rectangle is 54 cm and the area of
the square is 64 cm2, find the total area of the shaded rectangles.
Solution
Put the two shaded rectangles to form a long rectangle as shown:
Length of long rectangle =
1
1
2
 perimeter of unshaded rectangle
= 2  54 cm
= 27 cm
Breadth of long rectangle = length of square = 8 cm
 total area of shaded rectangles = 27 cm  8 cm = 216 cm2
22.
Find the value of √2 + √2 + √2+. . .
Solution
Let x = √2 + √2 + √2+. . .
Then x2 = 2 + √2 + √2+. . .
x2 = 2 + x
x2  x  2 = 0
(x + 1)(x  2) = 0
x = 2 or 1 (rejected because x is the positive square root)
 √2 + √2 + √2+. . . = 2
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SASMO 2015, Secondary 2 Contest
23.
How many rectangles are there in a 6  6 square grid?
Solution
There is exactly one rectangle for every pair of distinct horizontal lines and
every pair of distinct vertical lines.
 total no. of rectangles in a 6  6 square grid
= no. of pairs of distinct horizontal lines  no. of pairs of distinct vertical
lines
7×6
7×6
= 2  2
= 21  21
= 441
24.
In the following cryptarithm, all the different letters stand for different digits.
Find the 5-digit number EFCBH.
+
E
A
B
C
D
E
F
G
B
F
C
B
H
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 E = 1.
In the thousands column, A + E = 10 + F or A + E + 1 = 10 + F, and since E
= 1, then F = A  9 or A  8.
Now A  9, so F = 0 or 1. But E = 1 already, so F = 0.
In the hundreds column, if B + F = 10 + C, then B = 10 + C is not possible.
If B + F + 1 = 10 + C, then B = 9 + C means B = 9 and C = 0, but F = 0
already.
This means that there is no carryover from B + F or B + F + 1.
But B + F  C as F = 0 and B  C, so B + F + 1 = C, i.e. there is carryover
from the tens column. Since F = 0, then B + 1 = C. --- (1)
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SASMO 2015, Secondary 2 Contest
This also means that in the thousands column, A + E = 10, i.e. A = 9 since E
= 1.
Suppose there is no carryover in the ones column, i.e. D + B = H.
Then C + G = 10 + B because there is carryover from the tens column.
From (1), C = B + 1, so B + 1 + G = 10 + B, i.e. G = 9. But A = 9 already.
So there is a carryover in the ones column, i.e. D + B = 10 + H. --- (2)
Then C + G + 1 = 10 + B, i.e. B + 1 + G + 1 = 10 + B, which gives G = 8.
From (2), D + B = 10 + H.
Since D, B  7 (as G = 8 and A = 9 already) and D  B, the maximum value of
D + B is 7 + 6 = 13, i.e. H  3.
If B = 7, from (1), C = 8, but G = 8 already.
If B = 6, then D = 7. But from (1), C = 7 also, a contradiction. So H  3.
But H 2 (as F = 0 and E = 1 already), so H = 2. Then D + B = 7 + 5 or 5 +
7.
If B = 7, from (1), C = 8, but G = 8 already.
So D = 7 and B = 5. From (1), C = 6.
Thus E = 1, F = 0, A = 9, G = 8, H = 2, D = 7, B = 5, C = 6:
9
+
1
1
5
1
6
7
1
0
8
5
0
6
5
2
 EFCBH = 9567 + 1085 = 10 652
25.
Find the last five digits of 152015.
Solution
Since the last 5 digits of a product ab depends only on the last 5 digits of a
and of b, then
15 = 00 015
00 015  15 = 00 225
00 225  15 = 03 375
03 375  15 = 50 625
Last 5 digits of 50 625  15 = 59 375
Last 5 digits of 59 375  15 = 90 625
Last 5 digits of 90 625  15 = 59 375
 the last 5 digits repeat with a period of 2, with the exception of the first 4
powers.
Since the index 2015 is odd, then the last 5 digits of 152015 are 59 375.
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