AMC10 2006 A

(C) First note that, in general, the sum of $n$ consecutive integers is $n$ times their median. If the sum is $15$, we have the following cases: if $n=2$, then the median is $7.5$ and the two integers are $7$ and $8$; if $n=3$, then the median is $5$ and the three integers are $4$, $5$, and $6$; if $n=5$, then the median is $3$ and the five integers are $1$, $2$, $3$, $4$, and $5$. Because the sum of four consecutive integers is even, $15$ cannot be written in such a manner. Also, the sum of more than five consecutive integers must be more than $1+2+3+4+5=15$. Hence there are $3$ sets satisfying the condition. Note: It can be shown that the number of sets of two or more consecutive positive integers having a sum of $k$ is equal to the number of odd positive divisors of $k$, excluding $1$.