Question

一个有限的三位整数数列具有如下性质：每一项的十位和个位数字分别是下一项的百位和十位数字；最后一项的十位和个位数字分别是第一项的百位和十位数字。例如，这样的数列可能以 $247, 475$ 和 $756$ 开头，并以 $824$ 结尾。设 $S$ 为数列中所有项的和。总是整除 $S$ 的最大质数是多少？

A finite sequence of three-digit integers has the property that the tens and units digits of each term are, respectively, the hundreds and tens digits of the next term, and the tens and units digits of the last term are, respectively, the hundreds and tens digits of the first term. For example, such a sequence might begin with terms $247, 475,$ and $756$ and end with the term $824.$ Let $S$ be the sum of all the terms in the sequence. What is the largest prime number that always divides $S?$

$3$

$7$

$13$

$37$

$43$

答案：D

知识点：位值整除性

难度评级：1920

解答：

在整个数列中，每个数字作为百位、十位、个位出现的次数相同。

若 $k$ 是所有项个位数字之和，则 $S = 111k = 3 \cdot 37 \cdot k$ ，所以 $S$ 总能被 $37$ 整除。

数列 $123, 231, 312$ 给出 $S = 666 = 2 \cdot 3^2 \cdot 37$ ，没有更大的质因数被强制整除，所以答案是 $37$ 。

所以正确答案是 D。

Each digit appears as a hundreds digit, a tens digit, and a units digit the same number of times across the sequence.

If $k$ is the sum of the units digits of all terms, then $S = 111k = 3 \cdot 37 \cdot k,$ so $S$ is always divisible by $37.$

The sequence $123, 231, 312$ gives $S = 666 = 2 \cdot 3^2 \cdot 37,$ which has no larger prime factor forced, so $37$ is the answer.

Thus, the correct answer is D.

2007 AMC 10A 第 22 题

答案：D

难度评级：1920

解答：

其他年份的第 22 题