Question

从 $1$ 到 $2006$ （含端点）之间随机选择六个互不相同的正整数。某一对整数的差是 $5$ 的倍数的概率是多少？

Six distinct positive integers are randomly chosen between $1$ and $2006,$ inclusive. What is the probability that some pair of these integers has a difference that is a multiple of $5?$

$\dfrac{1}{2}$

$\dfrac{3}{5}$

$\dfrac{2}{3}$

$\dfrac{4}{5}$

$1$

答案：E

知识点：抽屉原理模运算

难度评级：1510

解答：

按模 $5$ 的余数给整数分组。可能余数只有 $5$ 种，但有 $6$ 个整数，所以根据鸽巢原理，必有两个整数余数相同。

它们的差就是 $5$ 的倍数。这一定发生，所以概率为 $1$ 。

所以正确答案是 E。

Group the integers by their remainder modulo $5.$ There are only $5$ possible remainders but $6$ integers, so by the Pigeonhole Principle two share a remainder.

Their difference is then a multiple of $5.$ This always happens, so the probability is $1.$

Thus, the correct answer is E.

2006 AMC 10A 第 20 题

答案：E

难度评级：1510

解答：

其他年份的第 20 题