HMMT 十一月 2019 · THM 赛 · 第 8 题
HMMT November 2019 — THM Round — Problem 8
题目详情
- Omkar, Krit , Krit , and Krit are sharing x > 0 pints of soup for dinner. Omkar always takes 1 pint of 1 2 3 soup (unless the amount left is less than one pint, in which case he simply takes all the remaining soup). 1 1 1 Krit always takes of what is left, Krit always takes of what is left, and Krit always takes of what 1 2 3 6 5 4 is left. They take soup in the order of Omkar, Krit , Krit , Krit , and then cycle through this order until 1 2 3 no soup remains. Find all x for which everyone gets the same amount of soup.
解析
- Omkar, Krit , Krit , and Krit are sharing x > 0 pints of soup for dinner. Omkar always takes 1 pint of 1 2 3 soup (unless the amount left is less than one pint, in which case he simply takes all the remaining soup). 1 1 1 Krit always takes of what is left, Krit always takes of what is left, and Krit always takes of what 1 2 3 6 5 4 is left. They take soup in the order of Omkar, Krit , Krit , Krit , and then cycle through this order until 1 2 3 no soup remains. Find all x for which everyone gets the same amount of soup. Proposed by: Krit Boonsiriseth and Milan Haiman 49 Answer: 3 The main observation is that if x > 1 pints of soup are left, then in one round, Omkar gets 1 and each x − 1 x − 1 Krit gets , with soup left. Thus it is evident that each Krit gets the same amount of soup, n n 6 2 x which means it suffices to find x for which Omkar gets . 4 Omkar gets 1 for each cycle and then all the remaining soup when there is less than one pint remaining. The amount of soup becomes (after each cycle) x − 1 x − 3 x + 1 x → → → · · · → − 1 , n 2 4 2 x +1 x so if n is the number of cycles, then Omkar’s soup is n + − 1. Setting this equal to , we obtain n 2 4 n n n + 1 / 2 − 1 ( n − 1)2 + 1 x = = . n n − 2 1 / 4 − 1 / 2 2 − 1 x +1 This immediately implies n > 2. On the other hand, we necessarily have 0 ≤ − 1 ≤ 1, so n 2 n n +1 2 ≤ x + 1 ≤ 2 . But n n − 2 n n ( n − 1)2 + 2 ( n − 1)2 + 2 x + 1 = ≤ = 8 n, n − 2 n − 3 2 − 1 2 n So 2 ≤ 8 n = ⇒ n ≤ 5. Testing n = 3 , 4 , 5: 4 • For n = 3 we get x = 17 which is greater than 2 . 49 • For n = 4 we get x = which works. 3 129 5 • For n = 5 we get x = which is less than 2 . 7 49 We see that only n = 4 and x = works. 3