Question

一群孩子举行吃葡萄比赛。比赛结束时，冠军吃了 $n$ 颗葡萄，第 $k$ 名孩子吃了 $n + 2 - 2k$ 颗葡萄。比赛中吃掉的葡萄总数为 $2009$ 。求 $n$ 的最小可能值。

A group of children held a grape-eating contest. When the contest was over, the winner had eaten $n$ grapes, and the child in $k$ th place had eaten $n + 2 - 2k$ grapes. The total number of grapes eaten in the contest was $2009.$ Find the smallest possible value of $n.$

答案

答案：89

知识点：等差数列因数分类讨论

难度评级：2110

解答：

设共有 $c$ 个孩子。各人的葡萄数 $n$ ， $n - 2$ ， $\ldots$ ， $n + 2 - 2c$ 构成等差数列，所以总数等于 $c$ 乘以首末两项的平均数： $\begin{aligned} &c \cdot \frac{n + (n + 2 - 2c)}{2} \\ &= c(n + 1 - c) \\ &= 2009 = 7^2 \cdot 41. \end{aligned}$ 因此 $c \mid 2009$ ，且 $n = \frac{2009}{c} + c - 1$ 。

最后一名孩子吃了 $n + 2 - 2c = \frac{2009}{c} + 1 - c \ge 0$ 颗葡萄，所以 $c(c - 1) \le 2009$ ，排除 $c = 49$ 、 $287$ 和 $2009$ 。剩余因数给出：当 $c = 1$ 时， $n = 2009$ ；当 $c = 7$ 时， $n = 287 + 6 = 293$ ；当 $c = 41$ 时， $n = 49 + 40 = 89$ 。

最小可能值为 $n = 89$ 。

Let $c$ be the number of children. The grape counts $n,$ $n - 2,$ $\ldots,$ $n + 2 - 2c$ form an arithmetic sequence, so the total is $c$ times the average of the first and last terms: $\begin{aligned} &c \cdot \frac{n + (n + 2 - 2c)}{2} \\ &= c(n + 1 - c) \\ &= 2009 = 7^2 \cdot 41. \end{aligned}$ Thus $c \mid 2009$ and $n = \frac{2009}{c} + c - 1.$

The last-place child ate $n + 2 - 2c = \frac{2009}{c} + 1 - c \ge 0$ grapes, which forces $c(c - 1) \le 2009,$ ruling out $c = 49,$ $287,$ and $2009.$ The remaining divisors give $n = 2009$ for $c = 1,$ $n = 287 + 6 = 293$ for $c = 7,$ and $n = 49 + 40 = 89$ for $c = 41.$

The smallest possible value is $n = 89.$

2009 AIME II 第 4 题

答案：89

难度评级：2110

解答：

其他年份的第 4 题