2006 AMC 10A 第 20 题

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

考虑这些整数除以

5

的余数。

Consider the remainders when the integers are divided by

5

6

个整数只有

5

种可能余数。

There are only

5

possible remainders for

6

integers

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.

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