2018IMAS Second Round Upper Primary Division Full Solutions

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Solution Key to Second Round of IMAS 2018/2019

Upper Primary Division

1. How many different prime divisors does

(2019−2018) (2019× −2017)× ×L (2019−2012) (2019× −2011) have? (A)2 (B)3 (C)4 (D)5 (E)8 【Solution】 (2019 2018) (2019 2017) (2019 2012) (2019 2011) 1 2 3 4 5 6 7 8 2 3 (2 2) 5 (2 3) 7 (2 2 2 2) − × − × × − × − = × × × × × × × = × × × × × × × × × × × L

There are 4 different divisors 2, 3, 5, 7. Hence (C).

Answer:(C) 2. Four students went mountain climbing together. They spent $50 in total on bus fare. While at the mountain top, each of them spent $5 on beverages. What is the average expense for each student?

(A)$12.5 (B)$13.75 (C)$17.5 (D)$30 (E)$55 【Solution 1】

Totally they spent $ 50+ × =5 4 70, so on average each spent $ 70÷ =4 17.5. 【Solution 2】

Average bus fare is $ 50÷ =4 12.5, so on average each spent $12.5 5 17.5+ = in total. Hence (C).

Answer:(C) 3. In the figure below, the side length of the square ABCD is 8 cm and the radius of

the sector AEF is 3 cm. What is the area, in cm2, of the shaded region? (Take π as

3.14, and round off to TWO decimal places)

(A)16.94 (B)19.38 (C)24.38 (D)26.94 (E)31.07 【Solution】

As in the figure, the area of the shaded area equals area of the square minus area of the sector and two right triangles. SinceBE = − =8 3 5cm. Thus the shaded area is

2 1 1 8 8 3.14 3 2 8 5 16.94 4 2 × − × × − × × × ≈ cm2. Hence (A). Answer:(A) A C B F E D

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4. In the figure below, a circular cone and a cube are shown. If a solid shape is formed by attaching the two shapes together, what is the least surface area, in cm2, of the resulting shape? (Take π as 3.14,, and round off to ONE decimal place)

(A)785.5 (B)942.5 (C)1000.5 (D)1021.0 (E)1099.5 【Solution】

To make the surface area as small as possible, the overlapping area should be as large as possible. For the circular cone, if the side is glued with the cube, the total surface area is not reduced. Hence the bottom of the cone should be glued to the cube. Since

the diameter of the bottom of the cone is 5 2 10× = cm, but side length of the cube is

12 cm, which is larger, the whole bottom face of the cone can be glued to part of a face of the cube. The total surface area of the resulted shape is at least

1

12 12 6 3.14 5 2 10 3.14 5 5 942.5

2

× × + × × × × − × × = cm2. Hence (B).

Answer:(B)

5. Given four distinct non-zero digits a, b, c and d, if ab+cd =dc+ba, then this expression is called a palindrome expression and the sum of the two numbers

ab+cd is called a palindrome sum. For example, 53 46+ =64 35+ =99. What is the minimum possible value of a palindrome sum?

(A)22 (B)33 (C)44 (D)55 (E)99 【Solution】

Since ab+cd =dc+ba, it is known that 10(a+ + +c) (b d) 10(= b+d)+ +(a c),

thus a+ = +c b d . The least number that can be represented as the sum of two

different positive numbers in two different ways is 5 1 4= + = +2 3, thus the least

palindrome sum is 55, for example, 12+43=34+21 55= . Hence (D).

Answer:(D)

6. There are a total of 40 students in a class. 23 of them are able to ride bikes, 33 of them are able to swim and 5 of them are unable to do either. How many students in this class are able to ride bikes but are not able to swim?

【Solution】

The number of students who are able to do at least one sport is 40 5− =35. Then the

number of students who can do both sports is 23 33 35+ − =21. The number of

students who can ride but not swim is then 23 21 2− = students.

Answer:2 students

10 cm

5 cm

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7. One day, Adam drove from Town A to Town B at a speed of 60 km/h. After an hour, the car stopped because of a breakdown, and because of this, Adam immediately called Bob for help. Bob then drove from A along the same route at a speed of 80 km/h. When Bob met Adam, he towed Adam’s car to B, at a speed of 40 km/h. The distance between A and B is 180 km. How long did Adam spend travelling for the whole trip?

【Solution】

It took Bob 60 80 3

4

÷ = hours to arrive at the place where Adam stopped. Then it

took (180 60 1)− × ÷40=3 hours for both cars to arrive at B. In total, it took Adam

3 3 1 3 4 4.75 4 4 + + = = hours from A to B. Answer:43 4.75 4 = hours 8. How many different prime numbers a are there such that a+20 and a+40

are also prime numbers? 【Solution 1】

Since 20 and 40 are both even numbers and a+20 and a+40 are primes, a≠2. Since the remainder of 20 divided by 3 is 2, the remainder of a divided by 3 can not be 1, otherwise a+20 is divisible by 3 and hence it is a composite number. This is a contradiction. Similarly, the remainder of a divided by 3 can not be 2, otherwise

40

a+ is a composite number. This is also a contradiction. So a is divisible by 3 and prime, which must be a =3, there is one possible value of a.

【Solution 2】

Regardless of the value of a, one of the three numbers a, a + 20, and a + 40 must be divisible by 3. From the given information, we known that a is a prime number, and a + 20 and a + 40 are also prime numbers. Therefore, a must be divisible by 3 and a prime number, it follows only a = 3, that is, there is one possible value of a.

Answer:1 9. Four identical chess pieces are to be placed into a 4 4× chess

board that is colored black and white alternately, as shown in the figure below. You can place at most one chess piece on each square. All chess pieces must be placed in squares of the same color and no two pieces are on the same row or on the same column. In how many different ways can the chess pieces be placed?

【Solution】

If all pieces were placed into black squares, there are two ways to put the pieces on the first and the third row, two ways to put the pieces on the second and the fourth row, totally there are 4 ways. Similarly, there are 4 ways to place all pieces into white squares. Hence there are 8 ways in total.

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10. The numbers 1, 2, 3, 4, 5 and 6 are written on the six faces of a unit cube without repetition. Each face contains one number and the sum of the numbers in every two opposite faces is 7. Put four such cubes side by side as shown in the figure below, such that sum of every two numbers of every two touched faces is 8. Find the number marked with “?” in the figure.

【Solution】

Since sum of numbers on opposite faces is 7, then 1 is opposite to 6, and 2 is opposite to 5, and 3 is opposite to 4. The right face of the first cube has number 3 or 4.

If it is 3, the left face of the second cube has number 5, right face has number 2, left face of the third cube has number 6, right face has 1, left face of the fourth cube has number 7, which is a contradiction.

If it is 4, the left face of the second cube has number 4, right face has number 3, left face of the third cube has number 5, right face has 2, left face of the fourth cube has number 6, right face has number 1, which is the place with mark “?”.

Answer:1

11. In the figure below, the side lengths of squares

ABCD and BEFG are 6 cm and 4 cm respectively

and triangle DFP is an isosceles right triangle. What is the area, in cm2, of triangle DFP?

【Solution】

Extend FG to intersect AD at point Q. Then triangle DQF is a right triangle, and

6 4 2

DQ= − = cm, QF = + =6 4 10cm.

By Pythagorean theorem DP2+PF2 =DF2 =DQ2 +QF2 =22 +102 =104. Since DP= PF, thenDP2 =52. Area of triangle DFP is1 1 2 26

DP PF× = ×2 DP = cm 2 . Answer:26 cm2 A C B E F D G P A C B E F D G Q P 1 5 ?

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12. A mouse starts from the top left-most unit square marked with “I”, follows a route to form the word “IMAS2019” by moving from one square to another square that share a common side. How many different routes of eight squares are there?

I M A S M A S 2 0 A S 2 0 1 S 2 0 1 9 0 1 9 【Solution】

In the following table, each square is filled with the number of routes reaching it. The number can be derived by recursion: each square is filled with sum of the numbers in adjacent squares with previous marks. From the table, it shows that the number of

different routes with length eight is 34 34+ =68.

1 1 1 1 1 2 3 4 4 1 3 6 10 14 1 4 10 20 34 4 14 34 Answer: 68 routes

13. If a and b are positive integers such that 1≤ < ≤a b 60 and a b× is divisible by 5. How many different ordered pairs of (a, b) are there?

【Solution】

Among positive integers less than or equal to 60, there are 12 numbers of them that are divisible by 5 and 48 numbers that are not. If both a and b are divisible by 5, one needs to choose 2 numbers among 12, so there are 12 11 66

2

× = pairs. If only one of

a, b is divisible by 5, and another is not divisible by 5, there are 12 48× =576 such

pairs. In total, there are 66+576=642 pairs.

Answer:642 pairs

14. In the figure, ABCD is a trapezoid, with side

AD that is parallel to BC, diagonals AC and

BD intersect at point O, and line OE is parallel to BC and intersects CD at point E. Now, extend OE to point F such that

OE =EF . If AD=6cm, BC=10 cm and the area of trapezoid ABCD is 64 cm2, what is the area, in cm2, of triangle ABF?

A C B E F D O

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【Solution 1】

Connect DF and CF, by properties of parallel lines one knows that SAOF =SDOF,

BOF COF

S =S , then SAOBF =SDOCF. E is midpoint of OF, then SDOC =SDCF, and

2 DOCF DOC S = S . (5 points) Since 6 3 10 5 ACD ABD ABC BCD S S AD S = S = BC = = △ △ △ △ , 3 24 5 3 ABD ABCD S = S = + △ cm 2 . (5 points) Since triangle DOA and triangle BOC are similar, we have 6 3

10 5

DO AD

OB = BC = = . (Or by common side theorem, 3

5 ACD ABC S DO OB = S = △ △ , then 3 5 ADO ABO S DO S = OB = △ △ . ) Hence 5 15 3 5 ABO ABD S = S = + △ △ cm 2 . (5 points)

As AD is parallel to BC, SABD =SACD, subtract SAOD on both sides one has

ABO DOC

S =S .

Hence SABF =SABO +SAOBF =SABO +SDOCF =SABO +2SDOC =3SABO =45cm2. (5 points)

【Solution 2】

As AD is parallel to BC, triangle AOD and COB are similar. Thus we have 6 3 10 5 DO AO AD OB = OC = BC = = and 2 2 2 6 9 ( ) 10 25 AOD COB S AD S = BC = = △ △ . (5 points)

Let the area of triangle AOD be 9x, then the area of triangle COB would be 25x. Then the areas of triangle AOB and COD are both 9 5 15

3

x× = x. So the area of trapezoid ABCD is 9x+25x+15x+15x=64x, i.e. x=1.Thus the areas of triangle AOB and COD are both 15 cm2 . (5 points)

Now, connect AE and BE. Since OE =EF, the areas of triangles DOE, AOE and AEF are all equal by the property of parallel lines. Thus the area of triangle AOF is twice of the area of triangle DOE. Similarly, the areas of triangles COE, BOE and BEF are all equal. Thus the area of triangle BOF is twice of the area of triangle COE. (5 points)

A C B E F D O A C B E F D O

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Observe that the area of triangle ABF is equal to the sum of areas of triangle ABO,

AOF and BOF, which is 15+ ×2 (SDOE +SCOE)= +15 2 1× 5=45 cm

2

. (5 points)

Answer: 45 cm2

15. A robot can generate a set of digit codes according to user’s reasonable instructions.Wayne gives out the following commands:

(1) Each code is a three-digit number (nonzero for the left-most digit). (2) Every two codes in the set have identical digits at no more than one

corresponding positions.

Find the maximum number of codes in a set the robot can generate. 【Solution 1】

From the given information, any two codes are different in at least two digits. Hence, the number of codes cannot exceed 90. Since, the hundred-digit can only be 9 digits from 1 to 9, the ten-digit can be 10 possible digits from 0 to 9, so the first two digits

in the 3-digit code number can form a total of 9 × 10 = 90 different two digits. If the

total number of codes is more than 90, then by pigeon-hole principle, at least two codes have the same hundred-digit and the same ten-digit, which is a contradiction!

(5 points)

Construct 90 codes as follows: the first two digits take 10 to 99, once for each; the

third digit equals the last digit of sum of first two digits. (5 points)

Next, we show that such set of codes satisfies the command. For any code abd,

a+ −b d is divisible by 10, thus each of a, b, d is determined by the other two. If two codes have identical digits at two corresponding places, their digits at the third place are also the same, showing that they are the same code. Thus every two codes

have identical digits at no more than one places. (10 points)

(5 points)

【Solution 2】

List down correctly all the 90 codes satisfying the command. (10 points, any error or

missing results in 0 points)

Prove that there are at most 90 codes. (10 points)

Answer:90 codes

【Note】

We also can construct 90 codes as follows: the first two digits take 10 to 99, once for each; the third digit satisfies that the sum of three digits is a multiple of 10. If two codes have one identical digits and the other one is different in the corresponding position, then the third digits are also different.

數據

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