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| − | == Problem ==
| + | #redirect [[2006 AMC 12A Problems/Problem 8]] |
| − | How many sets of two or more consecutive positive integers have a sum of 15?
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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>
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| − | == Solution ==
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| − | At a first glance, you should see that 7+8=15.
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| − | But are there three consecutive integers that add up to 15? Solve the equation
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| − | <math>n+n+1+n+2=15</math>, and you come up with n=4. 4+5+6=15.
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| − | Again solve the similar equation
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| − | <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.
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| − | So the answer is (C). 3
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| − | == See Also ==
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| − | {{AMC10 box|year=2006|ab=A|num-b=8|num-a=10}}
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| − | [[Category:Introductory Algebra Problems]] | |
