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 $n?$

$3$

$5$

$11$

$15$

$165$

答案：D

知识点：整除性最大公约数

难度评级：1480

小提示：

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

大提示：

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

解答：

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.

2003 AMC 10B 第 18 题

答案：D

难度评级：1480

小提示：

大提示：

解答：

其他年份的第 18 题