HMMT 十一月 2024 · THM 赛 · 第 5 题
HMMT November 2024 — THM Round — Problem 5
题目详情
- Alf, the alien from the 1980 s TV show, has a big appetite for the mineral apatite. However, he’s currently on a diet, so for each integer k ≥ 1 , he can eat exactly k pieces of apatite on day k . Additionally, if he eats apatite on day k , he cannot eat on any of days k + 1 , k + 2 , . . . , 2 k − 1 . Compute the maximum total number of pieces of apatite Alf could eat over days 1 , 2 , . . . , 99 , 100 .
解析
- Alf, the alien from the 1980s TV show, has a big appetite for the mineral apatite. However, he’s currently on a diet, so for each integer k ≥ 1, he can eat exactly k pieces of apatite on day k . Additionally, if he eats apatite on day k , he cannot eat on any of days k + 1, k + 2, . . . , 2 k − 1. Compute the maximum total number of pieces of apatite Alf could eat over days 1 , 2 , . . . , 99 , 100. Proposed by: Marin Hristov Hristov Answer: 197 Solution 1: If Alf doesn’t eat on day 100, he could have changed his diet so that he eats on all the same days except the last day is changed to 100. This attains strictly more apatite, and therefore an optimal diet must have Alf eating on day 100. Knowing this, Alf must not have eaten anything on days 51 , . . . , 99. Now, by the same logic, Alf must have eaten on day 50. Continuing the logic recursively gives that Alf must have eaten on days 100 , 50 , 25 , 12 , 6 , 3 , 1 . The sum of these numbers is 197 . Solution 2: The answer is 197 , achieved by Alf eating on days 1, 3, 6, 12, 25, 50, 100. We show that we could not do better. Let a > a > · · · > a be the days that Alf ate apatite. By problem’s condition, a ≥ 2 a for all i . 1 2 k i i +1 Thus, beginning with a ≤ 100, we deduce that 1 a 1 • a ≤ = 50, 2 2 a 2 • a ≤ = 25, 3 2 a 3 • a ≤ = 12, 4 2 a 4 • a ≤ = 6, 5 2 a 5 • a ≤ = 3, 6 2 a 6 • a ≤ = 1, 7 2 and hence k ≤ 7. Thus, a + · · · + a ≤ 100 + 50 + 25 + 12 + 6 + 3 + 1 = 197. 1 k Remark. The alternative answer 198193, which can be obtained if the contestant read 99 , 100 as the five-digit number 99100, was also accepted.