AMC10 2006 A

AMC10 2006 A · Q20

AMC10 2006 A · Q20. It mainly tests Pigeonhole principle, Remainders & modular arithmetic.

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?

从 1 到 2006（包含）中随机选取 6 个不同的正整数。求其中某些一对整数的差是 5 的倍数的概率是多少？

(A) $\frac{1}{2}$ $\frac{1}{2}$

(B) $\frac{3}{5}$ $\frac{3}{5}$

(D) $\frac{4}{5}$ $\frac{4}{5}$

(E) 1 1

Answer

Correct choice: (E)

正确答案：（E）

Solution

(E) Place each of the integers in a pile based on the remainder when the integer is divided by 5. Since there are only 5 piles but there are 6 integers, at least one of the piles must contain two or more integers. The difference of two integers in the same pile is divisible by 5. Hence the probability is 1. We have applied what is called the Pigeonhole Principle. This states that if you have more pigeons than boxes and you put each pigeon in a box, then at least one of the boxes must have more than one pigeon. In this problem the pigeons are integers and the boxes are piles.

（E）根据每个整数除以 5 的余数把这些整数分别放入不同的堆中。由于只有 5 堆但有 6 个整数，因此至少有一堆必须包含两个或更多整数。同一堆中的两个整数之差能被 5 整除。因此该概率为 1。我们使用了所谓的抽屉原理（鸽巢原理）。它说明：如果鸽子比盒子多，并且把每只鸽子都放进一个盒子里，那么至少有一个盒子必须包含不止一只鸽子。在本题中，鸽子是整数，盒子是这些堆。

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