Question

令 $N=123456789101112\dots4344$ 为把整数 $1$ 到 $44$ 依次写在一起形成的 $79$ 位数。 $N$ 除以 $45$ 的余数是多少？

Let $N=123456789101112\dots4344$ 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 $45?$

$1$

$4$

$9$

$18$

$44$

答案：C

知识点：模运算中国剩余定理数字

难度评级：1660

小提示：

分别求模 $5$ 和模 $9$ 的余数。

Find the remainder modulo $5$ and modulo $9$

大提示：

再把这两个条件合并成模 $45$ 的余数。

Combine them to get the remainder modulo $45$

解答：

要找除以 $45$ 的余数，必须分别找除以 $5$ 和 $9$ 的余数。除以 $5$ 的余数就是个位数字除以 $5$ 的余数，所以是 $4$ 。

要找除以 $9$ 的余数，通常找数字和。不过每个两位数除以 $9$ 的余数与它的数字和相同，所以可以直接把 $1$ 到 $44$ 的整数相加，因为每个整数与它自己的数字和同余。这个和为 $\dfrac{44\cdot45}{2}=990$ ，它是 $9$ 的倍数。因此 $N$ 是 $9$ 的倍数。

这个数是 $9$ 的倍数，并且除以 $5$ 的余数为 $4$ ，所以除以 $45$ 的余数为 $9$ 。

所以正确答案是 C。

To find the remainder when divided by $45,$ we must find the remainder when divided by $5$ and $9.$ The remainder when divided by $5$ is the remainder when the units digit is divided by $5,$ making it $4.$

To find the remainder when divided by $9,$ we usually find the sum of the digits. However, each double digit number has the same remainder when divided by $9$ as its digit sum, so we can just sum the integers from $1$ to $44,$ because each integer is congruent to its own digit sum. This sum is $\dfrac{44\cdot45}{2}=990,$ which is a multiple of $9.$ Thus, $N$ is a multiple of $9.$

Since it is a multiple of $9$ and has a remainder of $4$ when divided by $5,$ the remainder when divided by $45$ is $9.$

Thus, the correct answer is C .

2017 AMC 10B 第 23 题

答案：C

难度评级：1660

小提示：

大提示：

解答：

其他年份的第 23 题