2003 AMC 10B 第 18 题

Question

能整除下式的最大整数是多少？

\begin{aligned} &(n+1)(n+3)(n+5) \\ &\quad {}\cdot (n+7)(n+9) \end{aligned}

其中

n

为任意正偶数。

What is the largest integer that is a divisor of

\begin{aligned} &(n+1)(n+3)(n+5) \\ &\quad {}\cdot (n+7)(n+9) \end{aligned}

for all positive even integers

n?

3

5

11

15

165

当

n

为偶数时，五个因数是连续的五个奇数。

When

n

is even, the five factors are consecutive odd numbers

连续五个奇数中有一个是

3

的倍数，有一个是

5

的倍数；再检查没有更大的因数总是成立。

Among five consecutive odd numbers one is a multiple of

3

and one of

5;

check no larger factor always works

当

n

为偶数时，这五个因数是连续五个奇数。其中至少有一个能被

3

整除，且恰有一个能被

5

整除，所以乘积总能被

15

整除。

要证明不存在更大的固定公因数，比较以下三种情形：

\begin{aligned} n=2 &: 3\cdot5\cdot7\cdot9\cdot11,\\ n=10 &: 11\cdot13\cdot15\cdot17\cdot19,\\ n=12 &: 13\cdot15\cdot17\cdot19\cdot21 \end{aligned}\text{。}

这三个乘积的最大公因数恰为

15

，所以能整除所有情形的整数不可能更大。

所以正确答案是 D。

When

n

is even, the factors are five consecutive odd numbers. Among any five consecutive odd numbers, at least one is divisible by

3

and exactly one by

5,

so the product is always divisible by

15.

To prove that no larger fixed divisor is forced, compare three cases:

\begin{aligned} n=2 &: 3\cdot5\cdot7\cdot9\cdot11,\\ n=10 &: 11\cdot13\cdot15\cdot17\cdot19,\\ n=12 &: 13\cdot15\cdot17\cdot19\cdot21. \end{aligned}

The greatest common divisor of these three products is exactly

15,

so a divisor common to every case cannot be any larger.

Thus, the correct answer is D.

其他年份的第 18 题