# Difference between revisions of "2006 AMC 12A Problems/Problem 8"

The following problem is from both the 2006 AMC 12A #8 and 2008 AMC 10A #9, so both problems redirect to this page.

## Problem

How many sets of two or more consecutive positive integers have a sum of $15$? $\mathrm{(A) \ } 1\qquad \mathrm{(B) \ } 2\qquad \mathrm{(C) \ } 3\qquad \mathrm{(D) \ } 4\qquad \mathrm{(E) \ } 5$

## Solution

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 3, answer choice $\mathrm{(C) \ }$.

Question: (RealityWrites) Is it possible that the answer is 4, because 0+1+2+3+4+5 should technically count, right?

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 