2003 AMC 10B Problem 18

When $n$ is even, $n+1, n+3, n+5, n+7, n+9$ are five consecutive odd integers.

Why $15$ always divides. Consecutive odd numbers step through residues mod $3$ in the cycle $1, 0, 2, 1, 0, 2, \dots$ (each step adds $2$), so any three in a row contain a multiple of $3$. Likewise, adding $2$ repeatedly cycles through all residues mod $5$ in five steps, so five consecutive odds contain a multiple of $5$. Hence $3 \cdot 5 = 15$ divides the product for every even $n$.

Why nothing larger among the choices. The only larger choice is $165 = 3 \cdot 5 \cdot 11$, so we need one even $n$ for which $11$ fails. Take $n = 12$: the factors are $13, 15, 17, 19, 21$, none divisible by $11$. (Also $n = 4$ gives $5 \cdot 7 \cdot 9 \cdot 11 \cdot 13$ and $n = 2$ gives $3 \cdot 5 \cdot 7 \cdot 9 \cdot 11$, both divisible by $165$, which is exactly why $165$ is offered as a trap.)

The answer is $\boxed{\textbf{(D)}\ 15}$.

A run of $k$ consecutive terms of an arithmetic progression with difference coprime to $k$ covers every residue mod $k$, so one term is divisible by $k$; that is why $3$ and $5$ are guaranteed for five consecutive odds. Primes larger than the run length (here $7$, $11$) cannot be guaranteed. "For all $n$" problems have two halves: prove the divisor always works, then exhibit a single $n$ that kills any bigger candidate.

• Testing only $n = 2$ (product $10395 = 165 \cdot 63$) and concluding $165$, without checking a second value such as $n = 4$ or $n = 12$.
• Reasoning that five consecutive integers contain a multiple of $4$ or of $2$ and trying to include even factors; the factors here are all odd.