2017IMAS Second Round Junior Division Full Solutions

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Solution Key to Second Round of IMAS 2017/2018

Junior Division

1. Among all the expressions listed below, how many are negative numbers?

1

(1000 1)− , (1000−2)2, …, (1000−n)n, …, (1000−2018)2018. (A)509 (B)510 (C)1009 (D)1018 (E)1019

【Solution】

Since even powers of a real number are always non-negative, odd powers of a negative number are always negative and any powers of a positive number are always positive, (1000−n)n is negative for 1000− being negative and n being odd, thus, n n should be odd and over 1000. There are 2018 1000 509

2 −

= negative numbers. Answer:(A)

2. In convex quadrilateral ABCD, bisectors of DAB and ABC intersect at E, bisectors of ∠BCD and CDA intersect at F, as shown in the figure below. If∠AEB= ° , what is the angle measure, in degrees, of DFC80 ∠ ?

(A)80 (B)90 (C)100 (D)110 (E)Undetermined. 【Solution 1】 Since 180 ( ) 2 DAB ABC AEB ∠ + ∠ ∠ = ° − and 180 ( ) 2 CDA BCD DFC ∠ + ∠ ∠ = ° − , we have 360 ( ) 2

DAB ABC CDA BCD AEB DFC ∠ + ∠ + ∠ + ∠

∠ + ∠ = ° − .

And the sum of the inner angles of a quadrilateral is 360° ,

so 360∠AEB+ ∠DFC = ° −180° =180° and hence ∠DFC=180° − ° =80 100°. 【Solution 2】 180 ( ) 2 DAB ABC AEB ∠ + ∠ ∠ = ° − , 180 ( ) 2 CDA BCD DFC ∠ + ∠ ∠ = ° − ,

thus 2 (180∠DAB+ ∠ABC= × ° − ° =80 ) 200°. Since the sum of the inner angles of a quadrilateral is 360° , one has ∠CDA+ ∠BCD=360° −200° =160° , then

160 180 100 2 DFC ° ∠ = ° − = °. Answer:(C) A D E C B F

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3. Two numbers m and n, which may be equal, are taken from the set 1, 2, 3, 4, 5, 6, 7, 8 and 9. Which number below is not a possible value of 10(m+n)−mn?

(A)19 (B)55 (C)72 (D)79 (E)83

【Solution】

Since 10(m+n)−mn=100−(10−m)(10−n) , 9 10≥ − ≥ and 9 10m 1 ≥ − ≥ , n 1 10(m+n)−mn is less than 100, and its difference with 100 is (10-m)(10-n), which are the product of two digits. Observe that 100 19− =81 9 9= × , 100 55− =45= × , 100 72 28 4 79 5 − = = × , 100 79 21 3 7− = = × , but 100 83 17− = can not express as a product of two digits.

Answer:(E)

4. If a and b are real numbers, which of the following expressions below must be non-negative? (A) 2 2 a +b + + a b (B) 2018 2017 a +b (C) 4 4 2 2 1 a b +a b − (D) 3 3 2 2 2 a ba b +ab (E) 2 2 2 1 a b + ab+ 【Solution】 When 0a = , 1 2

b= − , the value of (A) is 1

4

− ; when a = , 10 b= − , the value of (B)

is − ; when 1 a = , 00 b= , the value of (C) is 1− ; when a = , 11 b= − , the value of

(D) is − ; but 4

a b

2 2

+

2

ab

+ =

1 (

ab

+

1)

2

0

.

Answer:(E)

5. The product of the sum and arithmetic mean of n integers is 2018. Which of the following statements below is true?

(A)Minimum of n is 1 (B)Minimum of n is 2 (C) Minimum of n is 1009 (D)Minimum of n is 2018 (E)No such n exists.

【Solution】

Denote by S as the sum of the n integers. Then S S 2018

n

× = , so S2 =2018n. Since 2018= ×2 1009 is a product of distinct primes, 2018 divides S and hence 20182 divides S . Thus 2018 divides n, i.e. 2 n≥2018 . On the other hand, take

2018

S = =n , S2 =2018 2018× =2018n is a solution to the problem.

Answer:(D)

6. Rotate an equilateral triangle inscribed in a circle 40 degrees clockwise and counter-clockwise, as shown in the figure below. How many triangles are there in the figure?

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【Solution】

(i) There are 9 triangles of same size but in different positions as the shaded triangle in the figure below.

(ii) There are 9 triangles of same size but in different positions as the shaded triangle in the figure below.

(iii) There are 9 triangles of same size but in different positions as the shaded triangle in the figure below.

(iv) There are 3 triangles of same size but in different positions as the shaded triangle in the figure below.

Totally there are 9+ + + =9 9 3 30 triangles.

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7. Consider a four-digit number abcd where a and d are both non-zero. If the last two digits in the sum of abcd and dcba are 58, what is the maximum possible value of abcd?

【Solution】

If a and b are not both 9, then abcd ≤9899. If a = = , then b 9 a+ =d 18 and 1 15

b+ + = . Thus, c d = and 9 c= , and the maximum value of 5 abcd is 9959. Answer: 9959

8. A rectangle is divided into 12 unit squares such that 10 are white and 2 are black, as shown in the figure below. To form a centrally symmetric picture by adding some white squares but no black squares, what is the least number of white squares needed?

【Solution】

Since no more black unit squares are added, the symmetric center of the whole picture is the symmetric center of the two black unit squares. 4 white unit squares are already symmetric to one another with respect to this center. 6 more white unit squares are needed to be symmetric to those 6 alone white unit squares, as in the figure to the right.

Answer: 6

9. A three-digit number is said to be "lucky" if it is divisible by 6 and by swapping its last two digits will give a number divisible by 6. How many "lucky" numbers are there?

【Solution】

Being divisible by 6 is equivalent to being divisible by both 2 and 3. Thus, the last two digits of a lucky number is even. A number is divisible by 3 if and only if the sum of all digits is divisible by 3. The remainder divided by 3 of the first number of the lucky number is determined by the last two digits. For any remainder, there are exactly 3 non-zero digits. Thus, the number of all lucky numbers is 5 5 3× × =75.

Answer: 75

Symmetric center

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10. Find the value of x such that both x and 2017−99 x are integers.

【Solution】

It is obvious that x should be an integer and 2017−99 x is a perfect square. Since 2017−99 x ≥ , 0 2017 21

99

x ≤ < . The possible values of x are 0, 1, 2, 3,

4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20. Similarly, 2017−99 x are respectively 2017, 1918, 1819, 1720, 1621, 1522, 1423, 1324, 1225, 1126, 1027, 928, 829, 730, 631, 532, 433, 334, 235, 136, 37. Only for x = , we get a perfect square 8

2017−99 x =1225. Thus x=82 =64.

Answer: 64

11. In the figure below, quadrilaterals ABCD and ABCE are both isosceles trapezoids, where AB//CE and BC//AD. If AC =DE, what is the measure, in degrees, of

ABC

∠ ?

【Solution】

Connect BE and BD. Since the diagonals of an isosceles trapezoid are equal, one knows that BE= AC=BD . By AC=DE , one knows that BDE is an equilateral triangle and ∠EBD= °60 . Now

2 180 60

EBD ABC ABE DBC

ABC BAC BCA

ABC

∠ = ∠ − ∠ − ∠ = ∠ − ∠ − ∠ = ∠ − ° = ° One gets ∠ABC=120°.

Answer: 120° A C B E D A C B E D

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12. Place 1 , 2 , 3 , …, 100 into several groups such that the sum of each group is not more than 10. Find the least number of groups needed to attain this kind of an arrangement?

【Solution】

Since 25+ 26 > + =5 5 10, 25 , 26 , 27 , …, 100 are arranged into different groups. So there are at least 76 groups. One the other hand, arrange

25− and 25 nn + into one group for n= , 2, …, 24 and each remaining 1 numbers into one group. Since ( 25+ +n 25−n)2 =50+2 252 −n2 <100, this gives a minimum number of 76 groups that is required to attain such arrangement.

Answer: 76

13. There is a sequence of five positive integers. Each number right after the first term is at least twice the number before it. If the sum of the five numbers is 2018, what is the least possible value of the last number?

【Solution】

Let x be the last number. Then the first four numbers are at most 16 x , 8 x , 4 x and 2 x . Thus 2018 16 8 4 2 x x x x x + + + + ≥ , i.e. 2018 16 104117 31 31 x≥ × = , so x is at least 1042. Taking the five numbers as 65, 130, 260, 521 and 1042 satisfies the requirement of the problem.

Answer: 1042

14. Let a, b, c and d be four positive integers such that b

a, c b , d c are simplified fractions and b c d

a + + is an integer. Prove that b c d ≥ −a 1. 【Solution】

From the condition of the problem, we know that b is coprime to a and c and c is coprime to b and d. (5 points)

Since b c d a + + is an integer, b c 2 (b c d) ac ac bc ad a + +b c = + + b is also an integer. So 2 ac

b is an integer. Since b is coprime to a and c, b=1. (5 points)

1 d

a + is then an integer, since both are reduced fractions, c a=c. (5 points)

1

d a

+

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15. In the figure below, ABC is a right isosceles triangle whereAB= AC. Let D be an exterior point such that BD= 2AD. Prove that ∠ADC+ ∠BDC = °45 .

【Solution】

Draw the right isosceles triangle EBD so that BED= °90 and E, C are on the opposite sides of BD, as shown in the figure below.

2 2

BD BE BE

BC = BA = BA and ∠EBA= ° − ∠45 ABD= ∠DBC implies that

~

EBA DBC

△ △ . (10 points)

So, one gets ∠BDC = ∠BEA. Thus

2

BD

DE= =DA implies ∠DEA= ∠DAE and

180 2 2(90 ) 2 2

EDA DEA DEA BEA BDC

∠ = ° − ∠ = ° − ∠ = ∠ = ∠ . (5 points) Then ∠ADC+ ∠BDC = ∠ADB+ ∠2 BDC = ∠ADB+ ∠EDA= °. 45 (5 points)

A C B D A C B E D

數據

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參考文獻

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