AMC12 2007 A

AMC12 2007 A · Q11

AMC12 2007 A · Q11. 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. For example, such a sequence might begin with the 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 factor that always divides $S$?

一个有限的三位整数序列具有如下性质：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，而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，这样的序列可能以 247、475、756 开头，并以 824 结尾。设 $S$ 为该序列所有项之和。问：总能整除 $S$ 的最大素因数是多少？

(A) 3 3

(B) 7 7

(D) 37 37

(E) 43 43

Answer

Correct choice: (D)

正确答案：（D）

Solution

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 $37\ \mathrm{(D)}$. To see that it need not be divisible by any larger prime, the sequence $123, 231, 312$ gives $S=666=2 \cdot 3^2 \cdot 37\Rightarrow \mathrm{\boxed{(D)~~37}}$.

某个给定数字作为某一项的百位、十位和个位出现的次数相同。设 $k$ 为所有项个位数字之和，则 $S=111k=3 \cdot 37k$，所以 $S$ 必须能被 $37\ \mathrm{(D)}$ 整除。为说明它不必被任何更大的素数整除，序列 $123, 231, 312$ 给出 $S=666=2 \cdot 3^2 \cdot 37\Rightarrow \mathrm{\boxed{(D)~~37}}$。

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