HMMT 十一月 2011 · THM 赛 · 第 2 题
HMMT November 2011 — THM Round — Problem 2
题目详情
- [ 3 ] In a game of Fish, R2 and R3 are each holding a positive number of cards so that they are collectivelyholding a total of 24 cards. Each player gives an integer estimate for the number of cards he is holding,
such that each estimate is an integer between 80% of his actual number of cards and 120% of his actualnumber of cards, inclusive. Find the smallest possible sum of the two estimates.
英文原题
A player Z is randomly chosen from the set of players who have at least as many cards as every
other player (i.e. Z has the most cards or is tied for having the most cards).
解析
英文解析
- [ 3 ] In a game of Fish, R2 and R3 are each holding a positive number of cards so that they are collectivelyholding a total of 24 cards. Each player gives an integer estimate for the number of cards he is holding,
such that each estimate is an integer between 80% of his actual number of cards and 120% of his actualnumber of cards, inclusive. Find the smallest possible sum of the two estimates.
Answer: 20 To minimize the sum, we want each player to say an estimate as small as possible–i.e.
an estimate as close to 80% of his actual number of cards as possible. We claim that the minimumpossible sum is 20.
First, this is achievable when R2 has 10 cards and estimates 8, and when R3 has 14 cards and estimates
Then, suppose that R2 has x cards and R3 has 24 − x . Then, the sum of their estimates is
⌈ ⌉ ⌈ ⌉ ⌈ ⌉ ⌈ ⌉
4 4 4 4 4
( x ) + (24 − x ) ≥ ( x ) + (24 − x ) ≥ (24) ≥ 20 .
5 5 5 5 5
Note: We use the fact that for all real numbers a, b , ⌈ a ⌉ + ⌈ b ⌉ ≥ ⌈ a + b ⌉ .