Difference between revisions of "2006 AMC 12A Problems/Problem 8"
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== Solution == | == Solution == | ||
| − | 1 | + | 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: |
| − | The correct answer is C (3) | + | *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 <math>\frac{1}{2}</math>. The only possibility is <math>\frac{15}{2}</math>, from which we get: | ||
| + | |||
| + | *7+8 = 15 | ||
| + | |||
| + | Thus, the correct answer is C (3). | ||
== See also == | == See also == | ||
| − | * [[2006 AMC 12A Problems]] | + | *[[2006 AMC 12A Problems]] |
| − | + | ||
| − | + | {{AMC box|year=2006|n=12A|num-b=7|num-a=9}} | |
[[Category:Introductory Algebra Problems]] | [[Category:Introductory Algebra Problems]] | ||
Revision as of 20:08, 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:
- 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
| {{{header}}} | ||
| Preceded by Problem 7 |
AMC 12A 2006 |
Followed by Problem 9 |
