How many [sets](https://artofproblemsolving.com/wiki/index.php/Set) of two or more consecutive positive integers have a sum of $15$?
$\textbf{(A) } 1\qquad \textbf{(B) } 2\qquad \textbf{(C) } 3\qquad \textbf{(D) } 4\qquad \textbf{(E) }  5$
Notice that if the consecutive positive integers have a sum of $15$, then their average (which could be a fraction) must be a divisor of $15$. If the number of integers in the list is odd, then the average must be either $1, 3,$ or $5$, and $1$ is clearly not possible. The other two possibilities both work:

$1 + 2 + 3 + 4 + 5 = 15$
$4 + 5 + 6 = 15$
If the number of integers in the list is even, then the average will have a $\frac{1}{2}$. The only possibility is $\frac{15}{2}$, from which we get:

$15 = 7 + 8$
Thus, the correct answer is $\boxed{\textbf{(C) }3}.$

Question: (RealityWrites) Is it possible that the answer is $4$, because $0+1+2+3+4+5$ should technically count, right?
Answer: (IMGROOT2) It isn't possible because the question asks for positive integers, and this means that negative integers or zero aren't allowed.
Note to readers: make sure to always read the problem VERY carefully before attempting; it could mean the difference of making the cutoff.