SW-ARML Practice 10-12-13 Solutions Individual Problems 1. Find the value of a2 + a4 + a6 + + a100 if a1, a2, a3, …, a100 is an arithmetic sequence with common difference 1, and a1 + a2 + a3 + + a100 = 137. 137 a1 + a2 + a3 + + a100 = (a2 – 1) + a2 + (a4 – 1) + a4 + + + (a100 – 1) + a100 137 50 93.5 = 2(a2 + a4 + a6 + + a100) – 50 a2 + a4 + a6 + + a100 = 2 2. 3. 200 What is the largest 2-digit prime factor of the integer ? 100 200 200! . Let p be a 2-digit prime such that 10 < p < 100. 100 100!100! If p 50, the p appears twice in the denominator, then p must appear three times in the numerator, or 3p < 200. Thus, the largest possible p is 61. (We don’t need to consider the case p < 50 as result.) Find the minimum value of 9 x 2 sin 2 x 4 for 0 < x < . x sin x 9 x 2 sin 2 x 4 9 x 2 sin 2 x 12 x sin x 4 (3x sin x 2) 12 12 12 and is equal to 12 = x sin x x sin x x sin x 2 when xsinx = (attainable by the intermediate value theorem and the fact xsinx = 0 when 3 π 2 π x = 0 and xsinx = > when x = ). 2 3 2 4. A point P is chosen at random in the interior of a unit square S. Let d(P) denote the distance 1 1 m from P to the closest side of S. The probability that d ( P ) is equal to , where m 5 3 n and n are relatively prime positive integers. Find m + n. 1 that have 3 the same center as the unit square and inside the square, and 3 inside the square with side length that have the same center 5 1 1 as the unit square d ( P ) . Since the area of the unit 5 3 Any point outside the square with side length 1 square is 1, the probability of a point P with 1 1 d ( P ) is the area of the shaded region, 5 3 2 2 9 1 56 3 1 which is the difference of the area of two squares: . Thus, the 25 9 225 5 3 answer is 281. 5. Positive numbers x, y, and z satisfy xyz = 1031 and log x (log yz) log y (log z) 168 . Find log x log y 2 2 (log z)2 . xyz = 1031 log xyz = 31 log x + log y + log z = 31. log x (log yz) log y (log z) 168 log x (log y) log x (log z) log y (log z) 168 (by using= log yz = log y log z ) Sine (a + b + c)2 = a2 + b2 + c2 + 2ab + 2ac + 2bc for all a, b and c, we have 2 2 log x log y (log z ) 2 = (log x + log y + log z)2 – 2[ log x (log y) log x (log z) log y (log z) ] = 312 2168 = 961 – 336 = 625. Hence, 6. log x log y 2 2 (log z )2 625 25 . Let N be the number of ordered pairs of nonempty sets A and B that have the following properties: A B = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12}, A B = , The number of elements of A is not an element of A, The number of elements of B is not an element of B. Find N. Let us partition the set {1, 2, 3, … , 12}into n numbers in A and 12 – n numbers in B, where 0 n 12, and n must be in B and 12 – n must be in A. In particular, n 6; otherwise, 6 would be in both A and B, but they are disjoint. Also, n 12; otherwise 0 A, an impossibility; Similarly, n 0. Hence, 1 n 5 or 7 n 11. 10 We now have ways of picking the additional numbers to be in A. Once the elements n 1 of A are chosen, the rest is in B. 2 So the answer is 11 10 10 10 10 10 9 8 7 6 N n1 210 252 1024 252 772 . n 0 n 1 5 n 1 2 3 4 5 7. Find the number of second-degree polynomials f(x) with integer coefficients and integer zeros for which f(0) = 2013. Let f(x) = a(x – r)(x – s) for some integers a, r, and s. Then, ars = 2013 = 31161. Case I: Both r and s are positive (and thus a). There are 33 = 27 ways to split up the prime factors among a, r, and s. However, r and s are indistiguishable. In one case, (a, r, s) = (2013, 1, 1), we have r = s. The other 26 cases are double counting, so there are 13, where r s. Case II: Either of r or s could be negative. i. When r s, there are four different possibilities (+, +), (+, ), (, +), and (, ). ii. Whenr = s, then (r, s) = (1, 1), (1, 1), and either (1, 1) or (1, 1). The total is 413 + 3 = 55. 8. Jar A contains four liters of a solution that is 45% acid. Jar B contains five liters of a solution that is 48% acid. Jar C contains one liter of a solution that is k% acid. From jar C, m liters of the solution is added to jar A, and the remainder of the solution in jar C is added n to jar B. At the end both jar A and jar B contain solutions that are 50% acid. Given that m and n are relatively prime positive integers, find k + m + n. One may first combine all three jars in to a single container. That container will have 10 liters of liquid, and it should be 50% acidic. Thus there must be 5 liters of acid. Jug A contained 45%4L = 1.8L of acid, and jug B 48%5L = 2.4Lof acid. Hence jar C contains 5 – 1.8 – 2.4 = 0.8L acid (out of 1 liter), so k = 80. m m m m 2 m m (80%) 1.8 4 (50%) 1.6 3.6 4 0.6 0.4 n n n n 3 n n So, m = 2 and n = 3, and k + m + n = 85. Then, 9. What is the product of the real roots of the equation x2 + 18x + 30 = 2 x2 18x 45 ? Squaring both sides will make it too difficult to handle. Instead, let y = x2 + 18x + 30. 3 Then the equation becomes y = 2 y 15 y2 = 4y + 60 y2 4y 60 = 0 (y – 10)(y + 6) = 0 y = 10 or y = 6 (extraneous) y = 10 x2 + 18x + 30 = 10 x2 + 18x + 20 = 0 the product of the real roots is 20. 10. Let an = 6n + 8n for all positive integers n. Determine the remainder upon dividing a83 by 49. 83 83 82 83 81 83 80 83 7 7 7 7 7 1 1 2 3 82 a83 = (7 – 1)83 + (7 + 1)83 = 83 83 83 83 783 782 781 780 7 1 1 2 3 82 83 83 83 83 = 2 783 781 779 73 7 2 4 80 82 83 All but the last term is divisible by 49, so the remainder is equal to the remainder of 2 7 82 83 when divided by 49. Since 2 7 = 2837 = (166)7 = (237 + 5)7 = 2349 + 35. 82 Hence, the remainder is 35. 4