Question

$7$ 位数 $\underline{74A52B1}$ 和 $\underline{326AB4C}$ 都是 $3$ 的倍数。下列哪一个可能是 $C$ 的值？

The $7$ -digit numbers $\underline{74A52B1}$ and $\underline{326AB4C}$ are each multiples of $3$ . Which of the following could be the value of $C$ ?

$1$

$2$

$3$

$5$

$8$

答案：A

知识点：整除性数字

难度评级：1300

解答：

为使 $\underline{74A52B1}$ 能被 $3$ 整除，数字和 $19+A+B$ 必须是 $3$ 的倍数。因此 $A+B$ 比某个 $3$ 的倍数小 $1$ 。

为使 $\underline{326AB4C}$ 能被 $3$ 整除，数字和 $15+A+B+C$ 必须是 $3$ 的倍数。因为 $15$ 已经能被 $3$ 整除，所以 $A+B+C$ 必须能被 $3$ 整除。

因此 $C$ 必须比某个 $3$ 的倍数大 $1$ 。选项中只有 $1$ 符合。

所以正确答案是 A。

For $\underline{74A52B1}$ to be divisible by $3$ , the digit sum $19+A+B$ must be a multiple of $3$ . Hence $A+B$ is $1$ less than a multiple of $3$ .

For $\underline{326AB4C}$ to be divisible by $3$ , the digit sum $15+A+B+C$ must be a multiple of $3$ . Since $15$ is already divisible by $3$ , $A+B+C$ must be divisible by $3$ .

Therefore $C$ must be $1$ more than a multiple of $3$ . Among the answer choices, only $1$ has that form.

Thus, A is the correct answer.

2014 AMC 8 第 21 题

答案：A

难度评级：1300

解答：

其他年份的第 21 题