Question

下式对所有正偶数都成立时，其最大公因数是多少？

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

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

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。

没有更大的固定因数总能整除这些乘积：例如比较时的乘积与等其他情形，可知最大公因数正是 $15$ 。 $\begin{aligned} n=2 &: 3\cdot5\cdot7\cdot9\cdot11,\\ n=10 &: 11\cdot13\cdot15\cdot17\cdot19,\\ n=12 &: 13\cdot15\cdot17\cdot19\cdot21. \end{aligned}$

所以正确答案是 D。

For even $n,$ 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

解答：

其他年份的第 18 题