2006 AMC 10A Problem 9

A run of $n$ consecutive integers has sum $n$ times its average, and that average is the middle term when $n$ is odd, or a number ending in $.5$ when $n$ is even. So we need $15/n$ to have the right form, and the run built around it must stay positive.

• $n = 2$: average $7.5$, run $7, 8$. Works.
• $n = 3$: average $5$, run $4, 5, 6$. Works.
• $n = 4$: average $3.75$, not a half-integer. Fails.
• $n = 5$: average $3$, run $1, 2, 3, 4, 5$. Works.
• $n = 6$: average $2.5$, run $0, 1, \dots, 5$, but $0$ is not positive. Fails.

Any longer run has average below $2.5$ and would start at or below $0$, so the search is complete: exactly $3$ sets.

The answer is $\boxed{\textbf{(C)}\ 3}$.

Consecutive integers are an arithmetic sequence, and "sum $=$ count $\times$ average" turns the search into a divisibility question on $15$: odd lengths need $n \mid 15$, even lengths need $n \mid 30$ with an odd quotient. Checking each possible length in order keeps the enumeration short and exhaustive.

With first term $a \ge 1$ and $n \ge 2$ terms, $n(2a + n - 1) = 30$. The two factors have opposite parity and the second exceeds the first, so try $n = 2, 3, 5$: they give $a = 7, 4, 1$. The next candidate $n = 6$ forces $a = 0$. Three solutions.

• Counting $0 + 1 + 2 + 3 + 4 + 5 = 15$ as a set even though $0$ is not positive, which gives $4$ (choice (D)).
• Stopping after finding $7 + 8$ and $4 + 5 + 6$ without testing the length-$5$ run.