AMC10 2007 A

AMC10 2007 A · Q22

AMC10 2007 A · Q22. It mainly tests Digit properties (sum of digits, divisibility tests), Number theory misc.

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. ... Let $S$ be the sum of all the terms in the sequence. What is the largest prime number that always divides $S$?

一个有限的三位整数序列具有如下性质：每个项的十位和个位数字分别作为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作为第一项的百位和十位数字。……令$S$为序列中所有项之和。总是整除$S$的最大素数是多少？

(A) 3 3

(B) 7 7

(D) 37 37

(E) 43 43

Answer

Correct choice: (D)

正确答案：（D）

Solution

Answer (D): A given digit appears as the hundreds digit, the tens digit, and the units digit of a term the same number of times. Let $k$ be the sum of the units digits in all the terms. Then $S = 111k = 3\cdot 37k$, so $S$ must be divisible by $37$. To see that $S$ need not be divisible by any larger prime, note that the sequence $123, 231, 312$ gives $S = 666 = 2\cdot 3^2 \cdot 37$.

答案（D）：某个数字作为百位、十位和个位出现的次数相同。设 $k$ 为所有项的个位数字之和。则 $S = 111k = 3\cdot 37k$，所以 $S$ 必须能被 $37$ 整除。要说明 $S$ 不必能被任何更大的素数整除，注意序列 $123, 231, 312$ 给出 $S = 666 = 2\cdot 3^2 \cdot 37$。

Topics