Difference between revisions of "2006 AMC 10A Problems/Problem 9"
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== See Also == | == See Also == | ||
*[[2006 AMC 10A Problems]] | *[[2006 AMC 10A Problems]] | ||
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| + | *[[2006 AMC 10A Problems/Problem 8|Previous Problem]] | ||
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| + | *[[2006 AMC 10A Problems/Problem 10|Next Problem]] | ||
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| + | [[Category:Introductory Algebra Problems]] | ||
Revision as of 15:43, 4 August 2006
Problem
How many sets of two or more consecutive positive integers have a sum of 15?
Solution
At a first glance, you should see that 7+8=15.
But are there three consecutive integers that add up to 15? Solve the equation
, and you come up with n=4. 4+5+6=15.
Again solve the similar equation
to determine if there are any four consecutive integers that add up to 15. This comes out with the non-integral solution 9/4. Further speculation shows that 1+2+3+4+5 = 15.
So the answer is (C). 3
