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 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?

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 (ProblemsAnswer KeyResources)
Preceded by
Problem 7
Followed by
Problem 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 (ProblemsAnswer KeyResources)
Preceded by
Problem 8
Followed by
Problem 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

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