Question

对所有正偶数 $n$ 下式总能被整除的最大整数是多少？ $\begin{aligned} &(n + 1)(n + 3)(n + 5) \\ &\quad {}\cdot (n + 7)(n + 9) \end{aligned}$

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：1530

解答：

当 $n$ 为偶数时，这五个因数是连续奇数。任意五个连续奇数中至少有一个能被 $3$ 整除，且有一个能被 $5$ 整除，所以乘积总能被 $15$ 整除。

没有更大的整数总是可行：当 $n = 20$ 和 $n = 10$ 时，乘积分别为 $21 \cdot 23 \cdot 25 \cdot 27 \cdot 29$ 和 $11 \cdot 13 \cdot 15 \cdot 17 \cdot 19$ ，它们的最大公因数是 $15$ 。

因此，正确答案是 D。

For even $n,$ the five factors are 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 $15.$

No larger divisor always works: the products for $n = 10$ and $n = 20$ are $11 \cdot 13 \cdot 15 \cdot 17 \cdot 19$ and $21 \cdot 23 \cdot 25 \cdot 27 \cdot 29,$ whose greatest common divisor is $15.$

Thus, the correct answer is D.

2003 AMC 12B 第 12 题

答案：D

难度评级：1530

解答：

其他年份的第 12 题