2007 AMC 12A 第 11 题

Question

一个由三位整数构成的有限数列具有如下性质：每一项的十位数字和个位数字分别是下一项的百位数字和十位数字，而最后一项的十位数字和个位数字分别是第一项的百位数字和十位数字。例如，这样的数列可以从

247

、

475

和

756

开始，并以

824

结束。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：1500

小提示：

在整个循环中，每个出现的数字作为百位、十位和个位的次数相同。

Across the whole cycle, each digit serves once as a hundreds digit, once as a tens digit, and once as a units digit

大提示：

总和是所有个位数字之和的

111

倍，而

111=3\cdot 37

。

The sum is

111

times the total of the units digits, and

111=3\cdot 37

解答：

由于循环性质，每个出现的数字在百位、十位和个位上出现的次数相同。

设所有项个位数字之和为

k

，则

S=100k+10k+k=111k

=3\cdot 37\cdot k

。

所以

S

总能被

37

整除。它不一定能被更大的数整除：数列

123,231,312

给出

S=666=2\cdot 3^2\cdot 37

。

因此，正确答案是 D。

Because of the cycling property, each digit that appears is used the same number of times in the hundreds, tens, and units places.

Let

k

be the sum of the units digits over all terms. Then

S=100k+10k+k=111k

=3\cdot 37\cdot k.

So

S

is always divisible by

37.

It need not be divisible by anything larger: the sequence

123,231,312

gives

S=666=2\cdot 3^2\cdot 37.

Thus, the correct answer is D.

2007 AMC 12A 第 11 题

答案：D

难度评级：1500

小提示：

大提示：

解答：

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