AMC10 2017 B

AMC10 2017 B · Q23

AMC10 2017 B · Q23. 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?

设$N=123456789101112\dots4344$是由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

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 + · · · + 9) + 10·1 + 10·2 + 10·3 + (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.

N除以5的余数显然是4。一个正整数除以9的余数为0当且仅当其数字和除以9的余数为0。N的数字和是4(0+1+2+⋯+9)+10·1+10·2+10·3+(4+0)+(4+1)+(4+2)+(4+3)+(4+4)=270，因此N是9的倍数。那么N-9也是9的倍数，且N-9的末位是5，因此也是5的倍数。于是N-9是45的倍数，N除以45的余数是9。

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