AMC10 2007 B

AMC10 2007 B · Q24

AMC10 2007 B · Q24. It mainly tests Divisibility & factors, GCD & LCM.

Let $n$ denote the smallest positive integer that is divisible by both 4 and 9, and whose base-10 representation consists of only 4's and 9's, with at least one of each. What are the last four digits of $n$?

设 $n$ 表示最小的既能被 4 和 9 整除，其十进制表示只由 4 和 9 组成且至少各有一个的正整数。$n$ 的最后四位数字是什么？

(A) 4444 4444

(B) 4494 4494

(D) 9444 9444

(E) 9944 9944

Answer

Correct choice: (C)

正确答案：（C）

Solution

Answer (C): Since $n$ is divisible by 9, the sum of the digits of $n$ must be a multiple of 9. At least one digit of $n$ is 4, so at least nine digits must be 4, and at least one digit must be 9. For $n$ to be divisible by 4, the last two digits of $n$ must each be 4. These conditions are satisfied by several ten-digit numbers, of which the smallest is 4,444,444,944.

答案（C）：由于$n$能被9整除，因此$n$的各位数字之和必须是9的倍数。$n$中至少有一位数字是4，所以至少有九位必须是4，并且至少有一位必须是9。要使$n$能被4整除，$n$的最后两位都必须是4。满足这些条件的十位数有多个，其中最小的是4,444,444,944。

Topics