2003 AMC 12B Problem 12

Below is the professionally curated solution for Problem 12 of the 2003 AMC 12B, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2003 AMC 12B solutions, or check the answer key.

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Concepts:divisibility greatest common divisor

Difficulty rating: 1530

12.

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 $n?$

$3$

$5$

$11$

$15$

$165$

Solution:

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 $n = 2$ and $n = 10$ are $3 \cdot 5 \cdot 7 \cdot 9 \cdot 11$ and $11 \cdot 13 \cdot 15 \cdot 17 \cdot 19,$ whose greatest common divisor is $15.$

Thus, the correct answer is D.

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