# 2006 AMC 12A Problems/Problem 8

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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 $15$?

$\textbf{(A) } 1\qquad \textbf{(B) } 2\qquad \textbf{(C) } 3\qquad \textbf{(D) } 4\qquad \textbf{(E) } 5$

## 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 $\boxed{\textbf{(C) }3}.$

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.

## See also

 2006 AMC 12A (Problems • Answer Key • Resources) Preceded byProblem 7 Followed byProblem 9 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 All AMC 12 Problems and Solutions
 2006 AMC 10A (Problems • Answer Key • Resources) Preceded byProblem 8 Followed byProblem 10 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 All AMC 10 Problems and Solutions

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

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