AMC12 2003 B

AMC12 2003 B · Q12

AMC12 2003 B · Q12. It mainly tests Divisibility & factors, GCD & LCM.

What is the largest integer that is a divisor of \[(n+1)(n+3)(n+5)(n+7)(n+9)\] for all positive even integers $n$?

对于所有正偶整数 $n$，使得 \[(n+1)(n+3)(n+5)(n+7)(n+9)\] 都能被整除的最大整数是多少？

(A) 3 3

(B) 5 5

(D) 15 15

(E) 165 165

Answer

Correct choice: (D)

正确答案：（D）

Solution

For all consecutive odd integers, one of every five is a multiple of 5 and one of every three is a multiple of 3. The answer is $3 \cdot 5 = 15$, so ${\boxed{\textbf{(D)15}}}$ is the correct answer.

在任意五个连续奇整数中，每五个里必有一个是 $5$ 的倍数，每三个里必有一个是 $3$ 的倍数。因此该乘积必能被 $3\cdot 5=15$ 整除，所以正确答案是 ${\boxed{\textbf{(D)15}}}$。

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