Difference between revisions of "2006 AMC 10A Problems/Problem 9"
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<math> \mathrm{(A) \ } 1\qquad \mathrm{(B) \ } 2\qquad \mathrm{(C) \ } 3\qquad \mathrm{(D) \ } 4\qquad \mathrm{(E) \ } 5 </math> | <math> \mathrm{(A) \ } 1\qquad \mathrm{(B) \ } 2\qquad \mathrm{(C) \ } 3\qquad \mathrm{(D) \ } 4\qquad \mathrm{(E) \ } 5 </math> | ||
== Solution == | == 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 | ||
| + | |||
| + | <math>n+n+1+n+2=15</math>, and you come up with n=4. 4+5+6=15. | ||
| + | |||
| + | Again solve the similar equation | ||
| + | |||
| + | <math>n+n+1+n+2+n+3=15</math> 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 | ||
| + | |||
== See Also == | == See Also == | ||
*[[2006 AMC 10A Problems]] | *[[2006 AMC 10A Problems]] | ||
Revision as of 16:41, 15 July 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
