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| − | == Problem ==
| + | #REDIRECT[[2002 AMC 12B Problems/Problem 7]] |
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| − | The product of three consecutive positive integers is <math>8</math> times their sum. What is the sum of the squares?
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| − | <math> \mathrm{(A) \ } 50\qquad \mathrm{(B) \ } 77\qquad \mathrm{(C) \ } 110\qquad \mathrm{(D) \ } 149\qquad \mathrm{(E) \ } 194 </math>
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| − | == Solution ==
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| − | === Solution 1 ===
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| − | Let the three consecutive positive integers be <math>a-1</math>, <math>a</math>, and <math>a+1</math>. So, <math>a(a-1)(a+1)=a^3-a=24a</math>. Rearranging and factoring, <math>a(a+5)(a-5)=0</math>, so <math>a=5</math>. Hence, the sum of the squares is <math>4^2+5^2+6^2=77\Longrightarrow\boxed{\mathrm{ (B) \ }}</math>.
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| − | === Solution 2 ===
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| − | Backsolving from the answers. We can easily note that the five given answers correspond to <math>(3^2+4^2+5^2)</math>, <math>(4^2+5^2+6^2)</math>, ..., <math>(7^2+8^2+9^2)</math>. We can easily check each of these five possibilities and pick the correct one.
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| − | (It is not necessary to discover the exact form of the given answers. The observation that the answer is between <math>50</math> and <math>194</math> is enough to bound the search to the five possibilities mentioned above.)
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| − | ==See Also==
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| − | {{AMC10 box|year=2002|ab=B|num-b=10|num-a=12}}
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| − | [[Category:Introductory Algebra Problems]] | |
