SW-ARML 10-24-10 Solutions Problem 1. A function f(x) is given by the table x f(x) 1 4 2 2 3 6 4 3 5 1 6 5 If an n 1 is a sequence given by a1 = 4 and ak+1 = f(ak) for all k ≥ 1, find a2010. Solution. Note that a1 = 4, a2 = 3, a3 = 6, a4 = 5, a5 = 1, a6 = 4, so the sequence repeats the five values {4, 3, 6, 5, 1} periodically. Since 2010 is divisible by 5, a2010 = 1. Problem 2. Let x, y, and z all be positive numbers, and let w be a positive number such that log x w 24, log y w 40, log xyz w 12 . Find log z w . Solution. log x w 24,log y w 40,log xyz w 12 w x24 y 40 ( xyz)12 . Therefore, w5 x120 , w3 y120 , w10 ( xyz )120 x120 y120 z120 w2 z120 w z 60 log z w 60 Problem 3. Find the product of the real roots of the equation x2 18x 30 2 x2 18x 45 ? Solution. Avoid squaring both sides. Instead, let y x 2 18 x 30 . Then, y 2 y 15 y2 – 4y – 60 = 0 y = 10 and y = –6. But, –6 6 2 6 15 , so y = 10 is the only possibility. Therefore, x 2 18 x 30 10 x 2 18 x 20 0 . The product of the two real roots is 20. Problem 4. The number 1557, 1227, and 1341 have something in common. Each is a four-digit number beginning with 1 that has exactly two identical digits. How many such four-digit numbers are there? Solution. If the two identical digits are both 1, then the number must be 11xy, 1x1y, or 1xy1, where x and y are distinct and neither is equal to 1. There are 9·8 possibilities in each case, with a total of 3·72 = 216 of such numbers. If the two identical digits are not 1, then the number must be 1xxy, 1xyx, or 1yxx, where x and y are distinct and neither is equal to 1. Again, there are 9·8 possibilities in each case, with a total of 3·72 = 216 of such numbers. Thus, the desired answer is 432. 12 22 32 Problem 5. Let a 1 3 5 10062 12 22 32 and b 2011 3 5 7 10062 . Find the integer closest to a b. 2013 12 22 12 32 22 10062 10052 10062 (k 1)2 k 2 2k 1 1 for all k. it . Since 1 3 5 2011 2013 2k 1 2k 1 10062 1006 502.75.. 503.25 and the closet integer to a – b is 503. follows that a b 1006 2013 Solution. a b Problem 6. How many zeros does the number 2010! 1005! 2 end? Solution. The number of zeros at the end of n! is the largest power of 10 that si a factor of n!. It is also the largest power of 5 that divides n! In general the largest power of the prime p in the prime factorization of n! is n ! n ! n ! p p 2 p 3 , where [x] is the greatest-integer function. Therefore, the largest power of 5 that divides 2010 2010 2010 2010 2010! 1005! 2 1005 1005 1005 1005 is 2 5 25 125 625 5 25 125 625 = 402 + 80 +16 + 3 – 2(201 + 40 + 8 + 1) = 1. Problem 7. Find the minimum value of 9 x 2 sin 2 x 4 for 0 < x < . x sin x 9 y2 4 4 Solution. Let y = xsin x > 0 and rewrite the given expression as 9 y . The Arithmetic Meany y 4 9y 4 4 y 9 y 6 9 y 12 (= 12 when y = Geometric Mean Inequality implies that 2 y y 2 9 x 2 sin 2 x 4 4 2 9 y , y ) . The minimum value of is 12 (when x sin x which can be attained by the 3 x sin x y 3 Intermediate Value Theorem., since 0 xsinx /2 on the interval [0, /2]. 200 What is the largest 2-digit prime factor of the integer ? 100 200 200! Solution. . We are looking for a prime 10 < p < 100. The factor of p appears twice in the 100 100!100! denominator, so we’ll need it to appear in the numerator three times, or 3p 200, implying p = 61. Problem 8. Problem 9. Let an n 1 be given by an 6n 8n all n ≥ 1. Find the remainder upon dividing a2010 by 49. Solution. 62010 82010 (7 1)2010 (7 1)2010 2(72010 72008 divisible by 49, so the remainder is 2. 72 1) . All but the last term are Problem 10. Let n be the smallest positive integer that is a multiple of 75 and has exactly 75 positive integer n divisors, including 1 and itself. Find . 75 Solution. In general, if the prime factorization of n is n p1a p2b p3c , then n has exactly (a 1)(b 1)(c 1) divisors, including 1 and itself. Sine 75 = 3·52 = 3·5·5 and 75 divides n, two of n’s prime factors are 3 and 5. To minimize n, we can introduce a third prime factor, 2. Also, we need the factor 5 to be raised the least power. n 243452 16 27 432 . therefore, 2434and 52 is the smallest possible. The desired answer 75 3 52