Question

一间暗室里的抽屉中有 $100$ 只红袜、 $80$ 只绿袜、 $60$ 只蓝袜和 $40$ 只黑袜。一个孩子从抽屉中逐只取出袜子，但看不到取出的袜子是什么颜色。至少必须取出多少只袜子，才能保证其中至少有 $10$ 双？（一双袜子是两只颜色相同的袜子。每只袜子至多计入一双。）

A drawer in a darkened room contains $100$ red socks, $80$ green socks, $60$ blue socks and $40$ black socks. A youngster selects socks one at a time from the drawer but is unable to see the color of the socks drawn. What is the smallest number of socks that must be selected to guarantee that the selection contains at least $10$ pairs? (A pair of socks is two socks of the same color. No sock may be counted in more than one pair.)

$21$

$23$

$24$

$30$

$50$

答案：B

知识点：抽屉原理奇偶性 worst case

难度评级：1850

小提示：

对每种颜色，所取袜子中至多有一只无法配对

For each color, at most one selected sock can remain unpaired

大提示：

利用袜子总数的奇偶性改进“四只未配对”的界，再构造一个只差一点的反例

Use the parity of the total to sharpen the four-unpaired bound, then construct a near-miss

解答：

取出 $23$ 只袜子时，数量为奇数的颜色种数也必须为奇数，所以至多为 $3$ 。因此，至多有 $3$ 只袜子未配对，余下至少 $20$ 只袜子可组成 $10$ 双。但是，取出 $22$ 只还不够：若四种颜色的数量分别为 $7,5,5,5$ ，则只能组成 $3+2+2+2=9$ 双。因此，最小数量为 $23$ 。

所以正确答案是 B。

With $23$ selected socks, the number of colors having an odd count must itself be odd, so it is at most $3.$ Thus at most $3$ socks are unpaired, leaving at least $20$ socks in $10$ pairs. But $22$ socks do not suffice: color counts $7,5,5,5$ produce only $3+2+2+2=9$ pairs. Therefore the minimum is $23.$

Thus the correct answer is B.

1986 AMC 12 第 17 题

答案：B

难度评级：1850

小提示：

大提示：

解答：

其他年份的第 17 题