PURPLE COMET MATH MEET April 2011 MIDDLE SCHOOL - SOLUTIONS. There are relatively prime positive integers m and n so that. = m n.

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1 PURPLE COMET MATH MEET April 0 MIDDLE SCHOOL - SOLUTIONS Copyright c Titu Andreescu and Jonathan Kane Problem There are relatively prime positive integers m and n so that Find m + n = m n. Answer: = = = 8 9 = 9 6 = m n. Thus, m = 9 and n = 6. The requested sum is = 4. Problem The diagram below shows a -sided figure made up of three congruent squares. The figure has total perimeter 60. Find its area.

2 Answer: 75 If the total perimeter is 60, each square has perimeter 0. Thus, each square has side length 5. Since each square then has area 5 = 5, the entire figure has area 5 = 75. Problem Find the sum of all two-digit integers which are both prime and are more than a multiple of 0. Answer: 5 The two-digit integers which are both prime and are more than a multiple of 0 are,, 4, 6, and 7. Their sum is = 5. Problem 4 Jerry buys a bottle of 50 pills. Using a standard hour clock, he sees that the clock reads exactly when he takes the first pill. If he takes one pill every five hours, what hour will the clock read when he takes the last pill in the bottle? Answer: Jerry will take the last pill 5 49 hours after he takes the first pill. Since 5 49 = 5 (44 + 5) 5 5 = 5 (mod ), the clock will read hour past, or when Jerry takes the last pill. Problem 5 Given that = 9 9 n, find n. Answer: 5 Add 0 9 to each side of the given equation to yield 6 = 9 9 n = 9 9 n = n = n. Thus, 6 = n and n = 5.

3 Problem 6 The following addition problem is not correct if the numbers are interpreted as base 0 numbers. In what number base is the problem correct? Answer: 4 Let the number base for this problem be b. The ones column in the problem adds to = 6 (mod b). Thus, b must be a divisor of 4. Since the problem includes the digit 8, b must be at least 9. This implies that either b = or b = 4. It is straightforward to check that only b = 4 works. Problem 7 When 8 is divided by 8, the result is ( m n ), where m and n are relatively prime positive integers. Find m n. Answer: = = 6 8 = 4 4 = = 47. ( ) ( 8 = 56 ) 9. The requested difference is

Problem 8 A square measuring 5 by 5 is partitioned into five rows of five congruent squares as shown below. The small squares are alternately colored black and white as shown.

4 Problem 8 A square measuring 5 by 5 is partitioned into five rows of five congruent squares as shown below. The small squares are alternately colored black and white as shown. Find the total area of the part colored black. Answer: 7 Each of the small squares has side length 5 5 =, so each has area = 9. There are shaded squares. Thus, the shaded area is 9 = 7. Problem 9 A jar contains one quarter red jelly beans and three quarters blue jelly beans. If Chris removes three quarters of the red jelly beans and one quarter of the blue jelly beans, what percent of the jelly beans remaining in the jar will be red? Answer: 0 At the start, 4 of the jelly beans are red, and 4 of them are removed, so at the end, the number of red jelly beans will be = = 6 of the original number of jelly beans. Similarly, at the end, the number of blue jelly beans will be = 6 6 = 9 6 of the original number of jelly beans. Therefore, at the end, the fraction of red jelly beans in the jar is = +9 = 0 = 0 percent. 4

5 Problem 0 Five rays OA, OB, OC, OD, and OE radiate in a clockwise order from O forming four non-overlapping angles such that EOD = COB, COB = BOA, while DOC = BOA. If E, O, A are collinear with O between A and E, what is the degree measure of DOB? Answer: 90 Let BOA = x. Then COB = x, DOC = x, and DOE = 4x. Thus x + x + x + 4x = 0x = 80 = x = 8. Then DOB = DOC + COB = x + x = 5x = 90. Problem How many numbers are there that appear both in the arithmetic sequence 0, 6,, 8, and the arithmetic sequence 0,,, 4,..., 000? Answer: 6 The first sequence includes every sixth integer from 0 up to 000. The second sequence includes every eleventh integer from 0 up to 000. The numbers contained in both sequences are, therefore, every sixty-sixth number from 0 to 000. That is, 0, 76, 4, There are = 6 such numbers. Problem When Troy writes his digits, his 0,, and 8 look the same right-side-up and upside-down as seen in the figure below. His 6 and 9 look like upside-down images of each other. None of his other digits look like digits when they are inverted. How many different five-digit numbers (which do not begin with the digit zero) can Troy write which read the same right-side-up and upside-down? Answer: 60 Each of the digits that Troy writes in a five-digit number must be one of the 5

6 five: 0,, 8, 6, 9. In addition, the numbers in positions 4 and 5 must be the up-side-down versions of the digits he writes in positions and, respectively. Also, the digit in position must be 0,, or 8 since it must look the same right-side-up and up-side-down. Finally, the digit in the first position cannot be 0 since no five-digit number can begin with the digit 0. Thus, there are 4 possible choices for the first digit, 5 choices for the second digit, three choices for the third digit, and the last two digits are determined by the choices for the first two digits. Therefore, there are 4 5 = 60 choices for five-digit numbers. Problem The diagram shows two equilateral triangles with side length 4 mounted on two adjacent sides of a square also with side length 4. The distance between the two vertices marked A and B can be written as m + n for two positive integers m and n. Find m + n. 6

7 Answer: Let the center of the square be point C. The distance from A to C is given by the height of the equilateral triangle, 4 =, plus half the length of the side of the square,. Thus, the distance from A to C is +. Triangle ACB is an isosceles right triangle, so its hypotenuse AB has length ( + ) = The requested sum is =. Problem 4 The five-digit number 0 is divisible by the sum of its digits = 5. Find the greatest five-digit number which is divisible by the sum of its digits. Answer: 9997 Let n be the greatest five digit number which is divisible by the sum of its digits, and suppose that n ends with the digit a. If n = a = a, then it would be divisible by a = 6 + a. Then 6 + a would also divide ( a) (6 + a) = But = Because 6 + a is between 6 and 45, and no integer between 6 and 45 has factors in common with the prime 67 or divides 4, n is not of the form a. If n = a = a for some digit a, then n would be divisible by a = 5 + a. It would follow that 5 + a would also divide ( a) (5 + a) = But = 5. Because 5 + a must be between 5 and 44, there is no way for it to divide 5 = 45 or have any factors in common with the prime, so n is not of the form a. If n = a = a for some digit a, then by arguing as above, this would mean the number 4 + a would divide ( a) (4 + a) = 9996 = The number 4 + a cannot have any factors in common with the prime 47, but it could divide 5. The only integer between 4 and 4 7

8 equal to a power of times a power of is 6. This happens when a =, and the number 9997 is divisible by the sum of its digits which is 6. Problem 5 In the diagram below, AB and CD are parallel, BXY = 45, DZY = 5, and XY = Y Z. What is the degree measure of Y XZ? Answer: 55 Extend XY until it intersects CD at X. Then XX Z = 45. Hence ZY X = = 0. This implies that XY Z = 80 0 = 70. Since XY Z is isosceles, Y XZ + 70 = 80 and Y XZ = 55. Problem 6 Let a and b be nonzero real numbers such that a + b = 0 and a + b =. What is the quotient when a + b is divided by ab? Answer: 509 Adding fractions gives a+b ab together gives 4a+4b ab a+b ab = 0 4 = 509. = 4 a+b ab = 0 and a+b ab =. Adding these two equations = 0 + = 0. Thus, the requested ratio is Problem 7 Find the number of ordered quadruples (a, b, c, d) where each of a, b, c, and d are (not necessarily distinct) elements of {,,, 4, 5, 6, 7} and 8

9 abc + 4abd + 5bcd is even. For example, (,, 5, ) and (,, 4, 6) satisfy the conditions. Answer: 07 The given sum is even exactly when the products abc and bcd are either both even or both odd. This happens either when the product bc is even and or when bc is odd and a and d are either both even or both odd. Of the 7 7 = 49 ways to assign values to b and d, 4 4 = 6 ways result in bd being odd, and 49 6 = ways result in bc being even. Similarly, there are 49 ways to assign values to a and d. Of these ways, 4 4 = 6 result in both a and d being odd, and = 9 result in both a and d being even. It follows that the total number of ways to assign values to a, b, c, and d so that abc + 4abd + 5bcd is even is (6 + 9) = 07. Problem 8 Find the positive integer n so that n is the perfect square closest to Answer: 0 The sum = 005 ( ) i = 8 = 00 0 = (0 )(0 + ) = 0. i= Hence the requested number is 0. Problem 9 How many ordered pairs of sets (A, B) have the properties:. A {,,, 4, 5, 6}. B {,, 4, 5, 6, 7, 8}. A B has exactly elements. 9

10 Answer: 70 The sets {,,, 4, 5, 6} and {,, 4, 5, 6, 7, 8} have five elements in common, so there are ( 5 ) = 0 ways to select three elements that the sets A and B have in common. Having selected the three shared elements, there are two choices of what to do with the element which can either be in or out of set A. Similarly, there are two choices for each of the elements 7 and 8 which can be in or out of set B. Of the elements in {,, 4, 5, 6}, three of the elements have been chosen to appear in both A and B, but there are three choices of what to do with each of the other two elements which can be placed either in the set A, in the set B, or left out of both sets. It follows that the number of ways to choose the sets A and B is 0 = 70. Problem 0 Let V be the set of vertices of a regular 5 sided polygon with center at point C. How many triangles have vertices in V and contain the point C in the interior of the triangle? Answer: 650 The number of triangles with vertices among the 5 vertices is ( 5 ) = 00. The triangles that do not contain the center C are exactly the triangles that are obtuse. Number the vertices of the regular 5 sided polygon in order from to 5. A triangle will have an obtuse angle at the point numbered if one of its other two vertices is numbered with a number less than, one is numbered with a number greater than, and the difference between these two numbers does not exceed. Thus, one of these vertices could be numbered k for k 0 while the other could be numbered m with m k +. That is, for each choice of k, there are k choices for m. It follows that the number of triangles with an obtuse angle at vertex is = 0 = 66. An obtuse triangle could have its obtuse angle at any of the 5 vertices, so it 0

11 follows that there are 5 66 = 650 obtuse triangles. Hence, the number of triangles that contain the point C is = 650.

MIDDLE SCHOOL - SOLUTIONS. is 1. = 3. Multiplying by 20n yields 35n + 24n + 20 = 60n, and, therefore, n = 20.

MIDDLE SCHOOL - SOLUTIONS. is 1. = 3. Multiplying by 20n yields 35n + 24n + 20 = 60n, and, therefore, n = 20. PURPLE COMET! MATH MEET April 208 MIDDLE SCHOOL - SOLUTIONS Copyright c Titu Andreescu and Jonathan Kane Problem Find n such that the mean of 7 4, 6 5, and n is. Answer: 20 For the mean of three numbers