Question

$5$ 位数 $\underline{2}$ $\underline{0}$ $\underline{1}$ $\underline{8}$ $\underline{U}$ 能被 $9$ 整除。这个数除以 $8$ 的余数是多少？

The $5$ -digit number $\underline{2}$ $\underline{0}$ $\underline{1}$ $\underline{8}$ $\underline{U}$ is divisible by $9.$ What is the remainder when this number is divided by $8?$

$1$

$3$

$5$

$6$

$7$

答案：B

知识点：整除性模运算

难度评级：960

视频讲解：

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文字解答：

注意，一个数能被 $9$ 整除，当且仅当它的各位数字之和也能被 $9$ 整除。

题中这个五位数的数字和为因为 $\underline{2}$ $\underline{0}$ $\underline{1}$ $\underline{8}$ $\underline{U}$ 能被 $9$ 整除， $11+U$ 也必须被 $9$ 整除。又因为 $U$ 是一位数字， $0\le U\le 9$ ，所以 $U$ 只能是 $7$ 。 $2+0+1+8+U= 11+U.$

现在这个五位数是二万零一百八十七。用除法计算得 $20187=2523\cdot 8 + 3$ ，所以余数是 $3$ 。

所以正确答案是 B。

Notice that a number is divisible by $9$ if and only if the sum of its digits is also divisible by $9.$

The sum of the digits of the 5-digit number in the problem is: $2+0+1+8+U= 11+U.$ As $\underline{2}$ $\underline{0}$ $\underline{1}$ $\underline{8}$ $\underline{U}$ is divisible by $9,$ $11+U$ must also be divisible by $9.$ Also, as $U$ is a digit, we know that $0\le U\le 9.$ This means that $U$ can only be $7.$

Now we know that the 5-digit number in question is 20187, and we want to find the remainder when we divide 20187 by 8. To solve this, simply use long division to see that $20187=2523\cdot 8 + 3.$ Therefore, the remainder is $3.$

Thus, the correct answer is B.

2018 AMC 8 第 7 题

答案：B

难度评级：960

视频讲解：

文字解答：

其他年份的第 7 题