2003 AMC 12B Problem 12

Attempt Problem 12 of the 2003 AMC 12B below, then check your answer against the professionally curated solution 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.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

12.

What is the largest integer that is a divisor of (n+1)(n+3)(n+5)(n+7)(n+9) \begin{aligned} &(n + 1)(n + 3)(n + 5) \\ &\quad {}\cdot (n + 7)(n + 9) \end{aligned} for all positive even integers n?n?

33

55

1111

1515

165165

Answer: D
Concepts:divisibilitygreatest common divisor
Difficulty rating: 1530
Solution:

For even n,n, the five factors are consecutive odd numbers. Among any five consecutive odd numbers, at least one is divisible by 33 and exactly one by 5,5, so the product is always divisible by 15.15.

No larger divisor always works: the products for n=10n = 10 and n=20n = 20 are 111315171911 \cdot 13 \cdot 15 \cdot 17 \cdot 19 and 2123252729,21 \cdot 23 \cdot 25 \cdot 27 \cdot 29, whose greatest common divisor is 15.15.

Thus, the correct answer is D.

← Problem 11#11
Full Exam

Problem 12 in Other Years