AMC12 2017 B

AMC12 2017 B · Q19

AMC12 2017 B · Q19. It mainly tests Remainders & modular arithmetic, Digit properties (sum of digits, divisibility tests).

Let $N = 123456789101112 \dots 4344$ be the 79-digit number that is formed by writing the integers from 1 to 44 in order, one after the other. What is the remainder when $N$ is divided by 45?

令 $N = 123456789101112 \dots 4344$ 为由 1 到 44 的整数依次写成的 79 位数。$N$ 除以 45 的余数是多少？

(A) 1 1

(B) 4 4

(D) 18 18

(E) 44 44

Answer

Correct choice: (C)

正确答案：（C）

Solution

Answer (C): The remainder when $N$ is divided by 5 is clearly 4. A positive integer is divisible by 9 if and only if the sum of its digits is divisible by 9. The sum of the digits of $N$ is $4(0+1+2+\cdots+9)+10\cdot1+10\cdot2+10\cdot3+(4+0)+(4+1)+(4+2)+(4+3)+(4+4)=270$, so $N$ must be a multiple of 9. Then $N-9$ must also be a multiple of 9, and the last digit of $N-9$ is 5, so it is also a multiple of 5. Thus $N-9$ is a multiple of 45, and $N$ leaves a remainder of 9 when divided by 45.

答案（C）：当 $N$ 除以 5 时余数显然是 4。一个正整数当且仅当其各位数字之和能被 9 整除时，才能被 9 整除。$N$ 的各位数字之和为 $4(0+1+2+\cdots+9)+10\cdot1+10\cdot2+10\cdot3+(4+0)+(4+1)+(4+2)+(4+3)+(4+4)=270$，因此 $N$ 必为 9 的倍数。于是 $N-9$ 也必为 9 的倍数，且 $N-9$ 的末位数字是 5，所以它也是 5 的倍数。因此 $N-9$ 是 45 的倍数，而 $N$ 除以 45 的余数是 9。

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