Question

求满足以下条件的五位正整数 $n$ 的个数：

• $n$ 能被 $5$ 整除，

• $n$ 的首位数字和末位数字相等，并且

• $n$ 的各位数字之和能被 $5$ 整除。

Find the number of five-digit positive integers, $n,$ that satisfy the following conditions:

• the number $n$ is divisible by $5,$

• the first and last digits of $n$ are equal, and

• the sum of the digits of $n$ is divisible by $5.$

答案

答案：200

知识点：数字整除性乘法原理

难度评级：2020

解答：

因为 $n$ 能被 $5$ 整除，所以末位数字为 $0$ 或 $5$ ；又因为首位数字等于末位数字且不能为 $0$ ，所以首末两位都必须是 $5$ 。这两位对数字和的贡献为 $10$ ，所以中间三位的和也必须是 $5$ 的倍数。

第二位和第三位可任意选择，共 $10 \cdot 10 = 100$ 种。无论它们的和是多少，第四位都必须落在模 $5$ 的某个指定余数类中，而 $0$ 到 $9$ 中每个余数类恰有 $2$ 个数字。因此总数为 $10 \cdot 10 \cdot 2 = 200$ 。

Since $n$ is divisible by $5,$ its last digit is $0$ or $5;$ since the first digit equals the last digit and cannot be $0,$ both are $5.$ The outer digits contribute $10$ to the digit sum, so the three middle digits must also sum to a multiple of $5.$

Choose the second and third digits freely, in $10 \cdot 10 = 100$ ways. Whatever their sum is, the fourth digit must land in a prescribed residue class modulo $5,$ and exactly $2$ of the digits $0$ through $9$ lie in each class. The count is $10 \cdot 10 \cdot 2 = 200.$

2013 AIME I 第 2 题

答案：200

难度评级：2020

解答：

其他年份的第 2 题