2006 AMC 12A Problems/Problem 8
Problem
How many sets of two or more consecutive positive integers have a sum of 
?
Solution
Notice that if the consecutive positive integers have a sum of 15, then their average (which could be a fraction) must be a factor 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 
. The only possibility is 
, from which we get:
- 7+8 = 15
Thus, the correct answer is C (3).
See also
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| Preceded by Problem 7 |
AMC 12A 2006 |
Followed by Problem 9 |
