AMC8 2018

AMC8 2018 · Q7

AMC8 2018 · Q7. It mainly tests Remainders & modular arithmetic, Digit properties (sum of digits, divisibility tests).

The 5-digit number $2\ 0\ 1\ 8\ 7$ is divisible by 9. What is the remainder when this number is divided by 8?

五位数 $2\ 0\ 1\ 8\ 7$ 能被9整除。这个数除以8的余数是多少？

(A) 1 1

(B) 3 3

(D) 6 6

(E) 7 7

Answer

Correct choice: (B)

正确答案：（B）

Solution

Answer (B): To be divisible by $9$, the sum of the digits must be divisible by $9$, so $2+0+1+8+U=11+U$ is divisible by $9$. Thus $U=7$, and the $5$-digit number is $20187$. Then because $20187=8\cdot2523+3$, the remainder is $3$.

答案（B）：要能被 $9$ 整除，各位数字之和必须能被 $9$ 整除，因此 $2+0+1+8+U=11+U$ 必须能被 $9$ 整除。于是 $U=7$，这个 $5$ 位数是 $20187$。又因为 $20187=8\cdot2523+3$，所以余数是 $3$。

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