2003 AMC 10B Problem 18

Question

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

When

n

is even, the five factors are consecutive odd numbers

Among five consecutive odd numbers one is a multiple of

3

and one of

5;

check no larger factor always works

When

n

is even, 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.

Problem 18 in other years