Question

有多少个 $4$ 位正整数，也就是 $1000$ 到 $9999$ 之间的整数，只含偶数数字且能被 $5$ 整除？

How many $4$ -digit positive integers (that is, integers between $1000$ and $9999,$ inclusive) having only even digits are divisible by $5?$

$80$

$100$

$125$

$200$

$500$

答案：B

知识点：乘法原理数字整除性

难度评级：980

视频讲解：

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文字解答：

末位只能是 $0$ ，这样才能被 $5$ 整除。千位可为 $2,4,6$ 或 $8$ ，共有四种；百位和十位各有 $5$ 种选择。因此共有 $4\cdot5\cdot5=100$ 个这样的整数。正确答案是 B。

The last digit must be $0$ , because the number is divisible by $5$ and all digits are even. The thousands digit can be $2,4,6,$ or $8$ , and each of the hundreds and tens digits has $5$ choices. Thus there are $4\cdot5\cdot5=100$ such integers. Thus, B is the correct answer.

2020 AMC 10A 第 6 题

答案：B

难度评级：980

视频讲解：

文字解答：

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