IMSO2006 Math Kunci
INTERNATIONL MATHENATICS AND SCIENCE OLYMPIAD
FOR PRIMARY SCHOOLS (IMSO) 2006
Mathematics Contest in Taiwan
1.
$475
7.
142857
2.
3
8.
(b)
3.
7:00 am on a Thursday
9.
10 cm2
4.
28°C
10.
C
5.
192
11.
2
6.
6
12.
12
INTERNATIONL MATHENATICS AND SCIENCE OLYMPIAD
FOR PRIMARY SCHOOLS (IMSO) 2006
Mathematics Contest in Taiwan
Name:111
1
School: 111111
Grade: 11111
number: 11111111
Short Answer: there are 12 questions, fill in the correct answers in the answer
sheet. Each correct answer is worth 10 points. Time limit: 90 minutes.
1. At a sale, Sam bought a radio valued at $500. If he received a discount of
5%, what was the price he paid?
$500×(1ů5%)=$475.
2. In how many ways can the number 2006 be expressed as the product of two
factors, each of them greater than 1?
2006=2×17×59=2×1003=17×118=59×34, so the answer is 3 ways.
3. Chien-Ming Wang is a Taiwanese pitcher for the New York Yankees, and is
scheduled to pitch in a game on Wednesday starting at 6:00 pm New York
time. If Taipei time is 13 hours ahead, when will a live broadcast on local
television begin?
The starting time in Taipei will be
6:00 pm on a Wednesday＋17 hours
= 6:00 am on a Thursday＋1 hour
= 7:00 am on a Thursday
4. In Pingtung, the average minimum temperature during a week was 30°C.
The minimum temperature on the first six days of that week were 27°C, 29°
C, 32°C, 33°C, 29°C and 32°C. What was the minimum temperature on the
last day of that week?
30°C×7ů27°Ců29°Ců32°Ců33°Ců29°Ců32°C =28°C
(Sol. 2, The different temperature between the average and the first six
days is (ů3°C)+(ů1°C)+(2°C)+(3°C)+(ů1°C)+(2°C) = 2°C, hence
the minimum temperature on the last day of that week is (30ů2)°C
=28°C.)
5. How many positive integers greater than 5000 can be formed with the digits
2, 4, 6, 7 and 8, if no digit is used more than once in a number?
i.
For four-digit numbers: There are only 3 choices, 6, 7 and 8, for
the leading digit. When you fix the leading digit, you will have 4
choices for second digit. As you have chosen the first two digits,
there are 3 choices for the third digit. Finally, you only have 2
choices for the last digit when you have decided the first three
digits. So there are 3×4×3×2=72 numbers satisfied the conditions.
ii.
For five-digit numbers: Since each five-digit number is greater
than 5000, there are 5×4×3×2×1=120 numbers satisfied the
conditions.
Hence there are 72+120=192 numbers satisfied the conditions.
6. The Student Council has 36 members, and the ratio of the number of boys
to the number of girls is 3:1. How many more girls should be added to the
Student Council so that the ratio of the number of boys to the number of
girls will be 9:5?
3
4
1
4
We have known that there are 36× =27 boys and 36× =9 girls. If we
want the ratio of the number of boys to the number of girls will be 9:5,
5
9
we need 27× =15 girls. So 15ů9=6 girls should be added.
7. The first digit of a 6-digit number is 1. We move this 1 so that it becomes
the last digit of the number, without disturbing the order of the other five
digits. For example, 123456 will become 234561. If the new number is
exactly 3 times the original number, what is the original number?
Set the original 6-digit number 1abcde. Thus we get 1abcde×3=abcde1.
Observe the unit digit of 3×e is 1. So e=7 and we get
1abcd7×3=abcd71. Since 3×7=21, the unit digit of 3×d is 7ů2=5. So
d=5 and 1abc57×3=abc571. Because of 3×5=15, the unit digit of 3×c
is 5ů1=4. So c=8 and 1ab857×3=ab8571. 3×8=24, so the unit digit of
3×b is 8ů2=6 and hence b=2. Thus 1a2857×3=a28571. Finally, we
can get a=4 because the unit digit of 3×a is 2. So the original 6-digit
number is 142857.
8. In the diagram, the net of a cube is shown on the right, and
four cubes are shown on the left. Which of the four cubes
may be folded from the net?
(a)
(c)
(b)
(d)
( b ), From the net of the cube that is shown on the right,
we can find the small square in the center of every face are unshaded.
9. Two polygons are overlapping such that 2 of the area of the smaller
3
polygon lies outside the larger polygon, and
8
9
of the area of the larger
polygon lies outside the smaller polygon. When not overlapping, the total
area of the two polygons is 120 cm2. What is the area of the overlapping
part of the polygons?
Since
2
3
of the smaller polygon lies outside the larger polygon and
8
9
of the larger polygon is lies outside the smaller polygon, the area of the
smaller polygon is 3 times of the area of the overlapping part and the
area of the larger polygon is 9 times of the area of the overlapping part.
Hence the sum of the areas of the two polygons (when not overlapping)
is 3+9=12 times of the area of the overlapping part. So the area of the
overlapping part is 120÷12=10 cm2.
10. In the diagram below, a square piece of paper is divided into 16 identical
boxes in a 4×4 array, each filled with a different letter. The piece of paper is
folded four times according to the following instructions
(1) After the first fold, the upper half covers the lower half.
(2) After the second fold, the left half covers the right half.
(3) After the third fold, the lower half covers the upper half.
(4) After the fourth fold, the right half covers the left half.
Which letter is in the box on the 10th layer from the top?
C
A
B
C
D
E
F
G
H
I
J
K
L
M
N
O
P
11. The Fibonacci numbers are 1, 1, 2, 3, 5, 6, 13, 21, 34, ... . Each term after
the first two is obtained by adding the last two terms. What is the greatest
value of the ratio of two neighboring terms of the Fibonacci numbers?
Observe that the sequence is an increasing sequence, so
Since an−1 + an = an+1 ,
an+1
a
≥1≥ n .
an
an+1
an+1 an−1 + an an−1
=
=
+1 ≤ 1+1 = 2 .
an
an
an
12. The sum of six positive integers is equal to their product. What is this
common value?
Assume a, b, c, d, e, f are positive intergers and a≦b≦c≦d≦e≦f .
Since a+b+c+d+e+f ≦6 f . If
(a)
a≧2, then a×b×c×d×e×f ≧32f＞6f. It is impossible.
(b)
a=1, b≧2, then a×b×c×d×e×f ≧16f＞6f. It is impossible.
(c)
a=b=1, c≧2, then a×b×c×d×e×f ≧8f＞6f. It is impossible.
(d)
a=b=c=1, d≧3, then a×b×c×d×e×f ≧9f＞6f. It is impossible.
a=b=c=1, d=2, and if
(i) e＞3, then a×b×c×d×e×f ＞6f, It is impossible.
(ii) e=3, then a+b+c+d+e+f=1+1+1+2+3+f=8+f=6f. It is
impossible.
(iii) e=2, then a+b+c+d+e+f=1+1+1+2+2+f=7+f=4f. It is
impossible.
(e) a=b=c=d=1, then a×b×c×d×e×f =ef=4+e+f.
Hence efůeůf+1=5=(eů1)(fů1)=1×5, we can obtain eů1=1,
fů1=5.
(f) a=b=c=d=e=1, then a×b×c×d×e×f =f≦6f. It is impossible.
Hence a=b=c=d=1, e=2, f=6, a+b+c+d+e+f =12.
FOR PRIMARY SCHOOLS (IMSO) 2006
Mathematics Contest in Taiwan
1.
$475
7.
142857
2.
3
8.
(b)
3.
7:00 am on a Thursday
9.
10 cm2
4.
28°C
10.
C
5.
192
11.
2
6.
6
12.
12
INTERNATIONL MATHENATICS AND SCIENCE OLYMPIAD
FOR PRIMARY SCHOOLS (IMSO) 2006
Mathematics Contest in Taiwan
Name:111
1
School: 111111
Grade: 11111
number: 11111111
Short Answer: there are 12 questions, fill in the correct answers in the answer
sheet. Each correct answer is worth 10 points. Time limit: 90 minutes.
1. At a sale, Sam bought a radio valued at $500. If he received a discount of
5%, what was the price he paid?
$500×(1ů5%)=$475.
2. In how many ways can the number 2006 be expressed as the product of two
factors, each of them greater than 1?
2006=2×17×59=2×1003=17×118=59×34, so the answer is 3 ways.
3. Chien-Ming Wang is a Taiwanese pitcher for the New York Yankees, and is
scheduled to pitch in a game on Wednesday starting at 6:00 pm New York
time. If Taipei time is 13 hours ahead, when will a live broadcast on local
television begin?
The starting time in Taipei will be
6:00 pm on a Wednesday＋17 hours
= 6:00 am on a Thursday＋1 hour
= 7:00 am on a Thursday
4. In Pingtung, the average minimum temperature during a week was 30°C.
The minimum temperature on the first six days of that week were 27°C, 29°
C, 32°C, 33°C, 29°C and 32°C. What was the minimum temperature on the
last day of that week?
30°C×7ů27°Ců29°Ců32°Ců33°Ců29°Ců32°C =28°C
(Sol. 2, The different temperature between the average and the first six
days is (ů3°C)+(ů1°C)+(2°C)+(3°C)+(ů1°C)+(2°C) = 2°C, hence
the minimum temperature on the last day of that week is (30ů2)°C
=28°C.)
5. How many positive integers greater than 5000 can be formed with the digits
2, 4, 6, 7 and 8, if no digit is used more than once in a number?
i.
For four-digit numbers: There are only 3 choices, 6, 7 and 8, for
the leading digit. When you fix the leading digit, you will have 4
choices for second digit. As you have chosen the first two digits,
there are 3 choices for the third digit. Finally, you only have 2
choices for the last digit when you have decided the first three
digits. So there are 3×4×3×2=72 numbers satisfied the conditions.
ii.
For five-digit numbers: Since each five-digit number is greater
than 5000, there are 5×4×3×2×1=120 numbers satisfied the
conditions.
Hence there are 72+120=192 numbers satisfied the conditions.
6. The Student Council has 36 members, and the ratio of the number of boys
to the number of girls is 3:1. How many more girls should be added to the
Student Council so that the ratio of the number of boys to the number of
girls will be 9:5?
3
4
1
4
We have known that there are 36× =27 boys and 36× =9 girls. If we
want the ratio of the number of boys to the number of girls will be 9:5,
5
9
we need 27× =15 girls. So 15ů9=6 girls should be added.
7. The first digit of a 6-digit number is 1. We move this 1 so that it becomes
the last digit of the number, without disturbing the order of the other five
digits. For example, 123456 will become 234561. If the new number is
exactly 3 times the original number, what is the original number?
Set the original 6-digit number 1abcde. Thus we get 1abcde×3=abcde1.
Observe the unit digit of 3×e is 1. So e=7 and we get
1abcd7×3=abcd71. Since 3×7=21, the unit digit of 3×d is 7ů2=5. So
d=5 and 1abc57×3=abc571. Because of 3×5=15, the unit digit of 3×c
is 5ů1=4. So c=8 and 1ab857×3=ab8571. 3×8=24, so the unit digit of
3×b is 8ů2=6 and hence b=2. Thus 1a2857×3=a28571. Finally, we
can get a=4 because the unit digit of 3×a is 2. So the original 6-digit
number is 142857.
8. In the diagram, the net of a cube is shown on the right, and
four cubes are shown on the left. Which of the four cubes
may be folded from the net?
(a)
(c)
(b)
(d)
( b ), From the net of the cube that is shown on the right,
we can find the small square in the center of every face are unshaded.
9. Two polygons are overlapping such that 2 of the area of the smaller
3
polygon lies outside the larger polygon, and
8
9
of the area of the larger
polygon lies outside the smaller polygon. When not overlapping, the total
area of the two polygons is 120 cm2. What is the area of the overlapping
part of the polygons?
Since
2
3
of the smaller polygon lies outside the larger polygon and
8
9
of the larger polygon is lies outside the smaller polygon, the area of the
smaller polygon is 3 times of the area of the overlapping part and the
area of the larger polygon is 9 times of the area of the overlapping part.
Hence the sum of the areas of the two polygons (when not overlapping)
is 3+9=12 times of the area of the overlapping part. So the area of the
overlapping part is 120÷12=10 cm2.
10. In the diagram below, a square piece of paper is divided into 16 identical
boxes in a 4×4 array, each filled with a different letter. The piece of paper is
folded four times according to the following instructions
(1) After the first fold, the upper half covers the lower half.
(2) After the second fold, the left half covers the right half.
(3) After the third fold, the lower half covers the upper half.
(4) After the fourth fold, the right half covers the left half.
Which letter is in the box on the 10th layer from the top?
C
A
B
C
D
E
F
G
H
I
J
K
L
M
N
O
P
11. The Fibonacci numbers are 1, 1, 2, 3, 5, 6, 13, 21, 34, ... . Each term after
the first two is obtained by adding the last two terms. What is the greatest
value of the ratio of two neighboring terms of the Fibonacci numbers?
Observe that the sequence is an increasing sequence, so
Since an−1 + an = an+1 ,
an+1
a
≥1≥ n .
an
an+1
an+1 an−1 + an an−1
=
=
+1 ≤ 1+1 = 2 .
an
an
an
12. The sum of six positive integers is equal to their product. What is this
common value?
Assume a, b, c, d, e, f are positive intergers and a≦b≦c≦d≦e≦f .
Since a+b+c+d+e+f ≦6 f . If
(a)
a≧2, then a×b×c×d×e×f ≧32f＞6f. It is impossible.
(b)
a=1, b≧2, then a×b×c×d×e×f ≧16f＞6f. It is impossible.
(c)
a=b=1, c≧2, then a×b×c×d×e×f ≧8f＞6f. It is impossible.
(d)
a=b=c=1, d≧3, then a×b×c×d×e×f ≧9f＞6f. It is impossible.
a=b=c=1, d=2, and if
(i) e＞3, then a×b×c×d×e×f ＞6f, It is impossible.
(ii) e=3, then a+b+c+d+e+f=1+1+1+2+3+f=8+f=6f. It is
impossible.
(iii) e=2, then a+b+c+d+e+f=1+1+1+2+2+f=7+f=4f. It is
impossible.
(e) a=b=c=d=1, then a×b×c×d×e×f =ef=4+e+f.
Hence efůeůf+1=5=(eů1)(fů1)=1×5, we can obtain eů1=1,
fů1=5.
(f) a=b=c=d=e=1, then a×b×c×d×e×f =f≦6f. It is impossible.
Hence a=b=c=d=1, e=2, f=6, a+b+c+d+e+f =12.