HMMT 十一月 2021 · THM 赛 · 第 7 题
HMMT November 2021 — THM Round — Problem 7
题目详情
- Let n be the answer to this problem. Box B initially contains n balls, and Box A contains half as many balls as Box B . After 80 balls are moved from Box A to Box B , the ratio of balls in Box A to Box B p is now , where p, q are positive integers with gcd( p, q ) = 1. Find 100 p + q . q
解析
- Let n be the answer to this problem. Box B initially contains n balls, and Box A contains half as many balls as Box B . After 80 balls are moved from Box A to Box B , the ratio of balls in Box A to Box B p is now , where p, q are positive integers with gcd( p, q ) = 1. Find 100 p + q . q Proposed by: Sheldon Kieren Tan Answer: 720 Solution: Originally, box A has n/ 2 balls and B has n balls. After moving, box A has n/ 2 − 80 balls and B has n + 80 balls. The answer to the problem is thus 100( n/ 2 − 80) + ( n + 80) 51 n − 80 · 99 ? = = n. gcd( n/ 2 − 80 , n + 80) gcd( n/ 2 − 80 , n + 80) Write d = gcd( n/ 2 − 80 , n +80) = gcd( n/ 2 − 80 , 240). Then the problem is equivalent nd = 51 n − 80 · 99 or (51 − d ) n = 80 · 99, with d | 240. Let’s try to solve this. Either 51 − d or n must be divisible by 5. In the latter case, where n is divisible by 5, we see that d must be as well. Therefore d is either 0 or 1 mod 5. If n is divisible by 4, then we know that d is even and thus 51 − d is odd. Therefore, since 16 | 80 · 99, n must be divisible by 16, meaning that d is divisible by 8. Alternatively, if n is not divisible by 4, then since 16 | 80 · 99, 51 − d must be divisible by 8, meaning that d is 3 mod 8. Therefore d is either 0 or 3 mod 8. Putting these results together, we find that d must either be 0, 11, 16, or 35 mod 40. Since d is a divisor of 240 and less than 51, we conclude that d is either 16 or 40. If d = 16, then 51 − d = 35, which does not divide 80 · 99. If d = 40, then we get n = 720, which ends up working.