Final IMSO 2009 Exploration
6th INTERNATIONAL MATHEMATICS AND SCIENCE OLYMPIAD FOR PRIMARY SCHOOL
IMSO 2009
EXPLORATION PROBLEMS
Yogyakarta, 8 – 14 November 2009
DIRECTORATE OF KINDERGARTEN AND PRIMARY DEVELOPMENT
DIRECTORATE GENERAL OF PRIMARY AND SECONDARY EDUCATION MANAGEMENT
MINISTRY OF NATIONAL EDUCATION
NAME_________________________________COUNTRY______________________________
Exploration
IMSO2009
Yogyakarta, 8-14 November
1. Find as many 5-digit numbers as possible, using all digits 1, 2, 3, 4, and 5,
satisfying the following properties:
The first and the fifth digits are not 1.
The first and the fifth digits are not 5.
The second digit is not 2.
The third digit is not 3.
The fourth digit is not 4.
Answer :
21453, 21534, 23154, 23514, 24153, 24513, 25134, 25143
31254, 31524, 31452, 34152, 34512, 35124, 35214, 35412,
41253, 41523, 41532, 43152, 43512, 45123, 45132, 45213.
Marking scheme:
22 ways (6 points) , 21 ways (4 points), 20 ways (3 points), 19 ways (2 points) 18
ways (1 points)
2. Consider a line l and a point C, which is not on l. If there are three different
points on l, say points X, Y, and Z, we can construct three different triangles,
namely triangles CXY, CXZ, and CYZ.
Questions:
a. What is the maximum number of triangles that can be constructed if there
are 5 different points on l ?
b. What is the maximum number of triangles that can be constructed if there
are 10 different points on l ?
c. What is the maximum number of triangles that can be constructed if there
are 51 different points on l ?
Answer & marking scheme :
a. 4 + 3 + 2 + 1 = 10
b. 9 + 8 + 7 + 6 + 5 + 4 + 3 + 2 + 1 = 45
c. 50 + 49 + . . . + 2 + 1 = 1275
Page 2 of 6
(1 point)
(2 point)
(3 point)
NAME_________________________________COUNTRY______________________________
Exploration
IMSO2009
Yogyakarta, 8-14 November
3. Muchlis has five different tasks: A, B, C, D, and E. Each task is done in one day.
Five friends help Muchlis. They are Ali, Benny, Charlie, Dennis, and Ethan. Each
friend helps him doing exactly one task.
Here is some information concerning how Muchlis is doing the five tasks.
Muchlis is doing task A a day after Ali helps him.
Muchlis is doing task B three days after Benny helps him.
On the fourth day Muchlis is doing task C or D.
Charlie helps Muchlis doing task A or D or E.
The one who helps Muchlis doing task E is Dennis or Benny.
Question: Complete the following table.
Day
Task
Helper
1
2
3
4
5
Answer:
Day
1
2
3
4
5
Task
9. C (remaining task)
3.A (from 1 and 3)
5.E (from 3 and 5)
8. D (from 4)
1.B (from 2 and 3)
Helper
4.Ali (from 1)
2.Benny (from 2)
6. Dennis (from 5)
7. Charlie (from 4)
10. Ethan (remaining helper)
Marking scheme:
2 cells (1 point), 3 cells (3 point), 4 cells (4 point), 5-9 cells (5 point), 10 cells (6 point)
4. Consider six boxes A, B, C, D, E, and F containing 10, 11, 12, 14, 19, and 24
marbles, respectively. We play a merging game as follows:
We choose two boxes and redistribute the marbles in the two boxes so that the
two boxes have the same number of marbles. This process is called a move and
we say that the two boxes are merged.
For example, in the first move we may merge B and E to get a new distribution:
10, 15, 12, 14, 15, and 24. We may not merge A and B since the total number of
marbles in these boxes is 21, which cannot be divided by 2.
Page 3 of 6
NAME_________________________________COUNTRY______________________________
Exploration
IMSO2009
Yogyakarta, 8-14 November
The game ends when no move can be done. In the following table, we merge B
and E, followed by A and C, followed by D and F, followed by A and D, and
finally, C and F. No more boxes can be merged after that.
Merged
boxes
Initial
condition
A
The number of marbles in box
B
C
D
E
F
10
11
12
14
19
24
Move 1.
B and E
10
15
12
14
15
24
Move 2.
A and C
11
15
11
14
15
24
Move 3.
D and F
11
15
11
19
15
19
Move 4
A and D
15
15
11
15
15
19
Move 5
C and F
15
15
15
15
15
15
Question: Given the initial condition, do the merging game with the least number
of moves.
Merged
boxes
Initial
condition
A
B
4
5
A
4
B
5
The number of marbles in box
C
D
E
F
G
H
32
35
The number of marbles in box
C
D
E
F
G
6
8
9
31
32
H
35
6
8
9
31
Move 1
Move 2
........
........
Merged
boxes
Initial
condition
Move 1.
B dan H
4
20
6
8
9
31
32
20
Move 2.
E dan F
4
20
6
8
20
20
32
20
Move 3.
D dan G
4
20
6
20
20
20
20
20
Move 4.
A dan C
5
20
5
20
20
20
20
20
Page 4 of 6
IMSO2009
NAME_________________________________COUNTRY______________________________
Exploration
Yogyakarta, 8-14 November
Marking scheme:
4 steps (6 point), 5 steps (4 point), 6 steps (2 point), Finish (1 point)
5. Consider the expression
.
We can replace each star (*) with either a plus sign (+) or a minus sign (–) so that
the result equals 25.
For example, a suitable replacement is 1 – 2 – 3 – 4 + 5 – 6 + 7 + 8 + 9 +10.
Find as many different ways as possible to replace each star (*) with either a plus
sign (+) or a minus sign (–) so that the result equals 25.
Solution:
If we replace all stars with (+) than the result is 55. If we replace the (+) sign
preceding a number A with (-) then the result is 55-2A. So to obtain 25, we must
replace the (+) signs preceding numbers whose sum is 15.
So this problem is equvalent to finding all ways to express 15 as a sum of at least
2 terms, where the terms are between 2 and 10.
2 terms: (5, 10), (6, 9), (7,8)
3 terms: (2,3,10), (2,4,9), (2,5,8), (2,6,7), (3,4,8), (3,5,7), (4, 5, 6)
4 terms: (2, 3, 4, 6)
Marking scheme:
11 ways (6 points)
10 ways (4 points)
9 ways (2 points)
8 ways (1 points)
Page 5 of 6
NAME_________________________________COUNTRY______________________________
Exploration
IMSO2009
Yogyakarta, 8-14 November
6. There are many kinds of nets of a rectangular box of size 2 x 1 x 1 units. An
example of such a net is given in the figure below.
A rectangular box of size 2 x 1 x 1 units and one of its nets
Question:
Construct as many nets of a rectangular box of size 2 x 1 x 1 units as possible on
a single sheet of 12 x 15 grid paper. Use color to distinguish different nets.
Solution:
Marking scheme:
16 nets (6 point)
15 nets (4 point)
14 nets (3 point)
13 nets (2 point)
12 nets (1 point)
End.
Page 6 of 6
IMSO 2009
EXPLORATION PROBLEMS
Yogyakarta, 8 – 14 November 2009
DIRECTORATE OF KINDERGARTEN AND PRIMARY DEVELOPMENT
DIRECTORATE GENERAL OF PRIMARY AND SECONDARY EDUCATION MANAGEMENT
MINISTRY OF NATIONAL EDUCATION
NAME_________________________________COUNTRY______________________________
Exploration
IMSO2009
Yogyakarta, 8-14 November
1. Find as many 5-digit numbers as possible, using all digits 1, 2, 3, 4, and 5,
satisfying the following properties:
The first and the fifth digits are not 1.
The first and the fifth digits are not 5.
The second digit is not 2.
The third digit is not 3.
The fourth digit is not 4.
Answer :
21453, 21534, 23154, 23514, 24153, 24513, 25134, 25143
31254, 31524, 31452, 34152, 34512, 35124, 35214, 35412,
41253, 41523, 41532, 43152, 43512, 45123, 45132, 45213.
Marking scheme:
22 ways (6 points) , 21 ways (4 points), 20 ways (3 points), 19 ways (2 points) 18
ways (1 points)
2. Consider a line l and a point C, which is not on l. If there are three different
points on l, say points X, Y, and Z, we can construct three different triangles,
namely triangles CXY, CXZ, and CYZ.
Questions:
a. What is the maximum number of triangles that can be constructed if there
are 5 different points on l ?
b. What is the maximum number of triangles that can be constructed if there
are 10 different points on l ?
c. What is the maximum number of triangles that can be constructed if there
are 51 different points on l ?
Answer & marking scheme :
a. 4 + 3 + 2 + 1 = 10
b. 9 + 8 + 7 + 6 + 5 + 4 + 3 + 2 + 1 = 45
c. 50 + 49 + . . . + 2 + 1 = 1275
Page 2 of 6
(1 point)
(2 point)
(3 point)
NAME_________________________________COUNTRY______________________________
Exploration
IMSO2009
Yogyakarta, 8-14 November
3. Muchlis has five different tasks: A, B, C, D, and E. Each task is done in one day.
Five friends help Muchlis. They are Ali, Benny, Charlie, Dennis, and Ethan. Each
friend helps him doing exactly one task.
Here is some information concerning how Muchlis is doing the five tasks.
Muchlis is doing task A a day after Ali helps him.
Muchlis is doing task B three days after Benny helps him.
On the fourth day Muchlis is doing task C or D.
Charlie helps Muchlis doing task A or D or E.
The one who helps Muchlis doing task E is Dennis or Benny.
Question: Complete the following table.
Day
Task
Helper
1
2
3
4
5
Answer:
Day
1
2
3
4
5
Task
9. C (remaining task)
3.A (from 1 and 3)
5.E (from 3 and 5)
8. D (from 4)
1.B (from 2 and 3)
Helper
4.Ali (from 1)
2.Benny (from 2)
6. Dennis (from 5)
7. Charlie (from 4)
10. Ethan (remaining helper)
Marking scheme:
2 cells (1 point), 3 cells (3 point), 4 cells (4 point), 5-9 cells (5 point), 10 cells (6 point)
4. Consider six boxes A, B, C, D, E, and F containing 10, 11, 12, 14, 19, and 24
marbles, respectively. We play a merging game as follows:
We choose two boxes and redistribute the marbles in the two boxes so that the
two boxes have the same number of marbles. This process is called a move and
we say that the two boxes are merged.
For example, in the first move we may merge B and E to get a new distribution:
10, 15, 12, 14, 15, and 24. We may not merge A and B since the total number of
marbles in these boxes is 21, which cannot be divided by 2.
Page 3 of 6
NAME_________________________________COUNTRY______________________________
Exploration
IMSO2009
Yogyakarta, 8-14 November
The game ends when no move can be done. In the following table, we merge B
and E, followed by A and C, followed by D and F, followed by A and D, and
finally, C and F. No more boxes can be merged after that.
Merged
boxes
Initial
condition
A
The number of marbles in box
B
C
D
E
F
10
11
12
14
19
24
Move 1.
B and E
10
15
12
14
15
24
Move 2.
A and C
11
15
11
14
15
24
Move 3.
D and F
11
15
11
19
15
19
Move 4
A and D
15
15
11
15
15
19
Move 5
C and F
15
15
15
15
15
15
Question: Given the initial condition, do the merging game with the least number
of moves.
Merged
boxes
Initial
condition
A
B
4
5
A
4
B
5
The number of marbles in box
C
D
E
F
G
H
32
35
The number of marbles in box
C
D
E
F
G
6
8
9
31
32
H
35
6
8
9
31
Move 1
Move 2
........
........
Merged
boxes
Initial
condition
Move 1.
B dan H
4
20
6
8
9
31
32
20
Move 2.
E dan F
4
20
6
8
20
20
32
20
Move 3.
D dan G
4
20
6
20
20
20
20
20
Move 4.
A dan C
5
20
5
20
20
20
20
20
Page 4 of 6
IMSO2009
NAME_________________________________COUNTRY______________________________
Exploration
Yogyakarta, 8-14 November
Marking scheme:
4 steps (6 point), 5 steps (4 point), 6 steps (2 point), Finish (1 point)
5. Consider the expression
.
We can replace each star (*) with either a plus sign (+) or a minus sign (–) so that
the result equals 25.
For example, a suitable replacement is 1 – 2 – 3 – 4 + 5 – 6 + 7 + 8 + 9 +10.
Find as many different ways as possible to replace each star (*) with either a plus
sign (+) or a minus sign (–) so that the result equals 25.
Solution:
If we replace all stars with (+) than the result is 55. If we replace the (+) sign
preceding a number A with (-) then the result is 55-2A. So to obtain 25, we must
replace the (+) signs preceding numbers whose sum is 15.
So this problem is equvalent to finding all ways to express 15 as a sum of at least
2 terms, where the terms are between 2 and 10.
2 terms: (5, 10), (6, 9), (7,8)
3 terms: (2,3,10), (2,4,9), (2,5,8), (2,6,7), (3,4,8), (3,5,7), (4, 5, 6)
4 terms: (2, 3, 4, 6)
Marking scheme:
11 ways (6 points)
10 ways (4 points)
9 ways (2 points)
8 ways (1 points)
Page 5 of 6
NAME_________________________________COUNTRY______________________________
Exploration
IMSO2009
Yogyakarta, 8-14 November
6. There are many kinds of nets of a rectangular box of size 2 x 1 x 1 units. An
example of such a net is given in the figure below.
A rectangular box of size 2 x 1 x 1 units and one of its nets
Question:
Construct as many nets of a rectangular box of size 2 x 1 x 1 units as possible on
a single sheet of 12 x 15 grid paper. Use color to distinguish different nets.
Solution:
Marking scheme:
16 nets (6 point)
15 nets (4 point)
14 nets (3 point)
13 nets (2 point)
12 nets (1 point)
End.
Page 6 of 6