Difference between revisions of "2006 AMC 12A Problems/Problem 8"
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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: | 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 | + | *<math>1 + 2 + 3 + 4 + 5 = 15</math> |
| − | *4+5+6 = 15 | + | *<math>4 + 5 + 6 = 15</math> |
If the number of integers in the list is even, then the average will have a <math>\frac{1}{2}</math>. The only possibility is <math>\frac{15}{2}</math>, from which we get: | If the number of integers in the list is even, then the average will have a <math>\frac{1}{2}</math>. The only possibility is <math>\frac{15}{2}</math>, from which we get: | ||
| − | *7+8 = 15 | + | *<math>7 + 8 = 15</math> |
Thus, the correct answer is C (3). | Thus, the correct answer is C (3). | ||
Revision as of 20:11, 30 January 2007
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:
If the number of integers in the list is even, then the average will have a 
. The only possibility is 
, from which we get:
Thus, the correct answer is C (3).
See also
| {{{header}}} | ||
| Preceded by Problem 7 |
AMC 12A 2006 |
Followed by Problem 9 |
