Question

一个罐中有 $n$ 个弹珠。每个弹珠不是红色就是蓝色，且每种颜色至少有 $7$ 个弹珠。从罐中不放回地随机抽取 $7$ 个弹珠时，恰好 $4$ 个为红色的概率等于恰好 $5$ 个为红色的概率。求使这件事可能的最小五个 $n$ 值之和。

An urn contains $n$ marbles. Each marble is either red or blue, and there are at least $7$ marbles of each color. When $7$ marbles are drawn randomly from the urn without replacement, the probability that exactly $4$ of them are red equals the probability that exactly $5$ of them are red. Find the sum of the five least values of $n$ for which this is possible.

答案

答案：190

知识点：无放回抽样组合丢番图方程

难度评级：2390

解答：

设红弹珠有 $r$ 个、蓝弹珠有 $b$ 个，且 $r, b \ge 7$ 。条件为 $\binom{r}{4}\binom{b}{3} = \binom{r}{5}\binom{b}{2}$ 。因为 $\binom{r}{5} = \binom{r}{4}\frac{r-4}{5}$ 且 $\binom{b}{3} = \binom{b}{2}\frac{b-2}{3}$ ，约去公共因子得 $\frac{b - 2}{3} = \frac{r - 4}{5},$ $b = \frac{3r - 2}{5}.$

因此 $r \equiv 4 \pmod 5$ ，而 $b \ge 7$ 要求 $3r - 2 \ge 35$ ，所以 $r \ge 14$ 。最小的五个选择为 $r = 14, 19, 24, 29, 34$ ，对应 $b = 8, 11, 14, 17, 20$ ，给出 $n = 22, 30, 38, 46, 54$ 。

和为 $22 + 30 + 38 + 46 + 54 = 190$ 。

Say there are $r$ red and $b$ blue marbles, $r, b \ge 7.$ The condition is $\binom{r}{4}\binom{b}{3} = \binom{r}{5}\binom{b}{2}.$ Since $\binom{r}{5} = \binom{r}{4}\frac{r-4}{5}$ and $\binom{b}{3} = \binom{b}{2}\frac{b-2}{3},$ cancelling gives $\frac{b - 2}{3} = \frac{r - 4}{5},$ that is, $b = \frac{3r - 2}{5}.$

So $r \equiv 4 \pmod 5,$ and $b \ge 7$ requires $3r - 2 \ge 35,$ so $r \ge 14.$ The five smallest choices are $r = 14, 19, 24, 29, 34$ with $b = 8, 11, 14, 17, 20,$ giving $n = 22, 30, 38, 46, 54.$

The sum is $22 + 30 + 38 + 46 + 54 = 190.$

2026 AIME II 第 5 题

答案：190

难度评级：2390

解答：

其他年份的第 5 题