Difference between revisions of "2002 AMC 10B Problems/Problem 11"

Latest revision as of 18:07, 28 July 2011

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-== Problem ==
+#REDIRECT[[2002 AMC 12B Problems/Problem 7]]
-The product of three consecutive positive integers is <math>8</math> times their sum. What is the sum of the squares?
-<math> \mathrm{(A) \ } 50\qquad \mathrm{(B) \ } 77\qquad \mathrm{(C) \ } 110\qquad \mathrm{(D) \ } 149\qquad \mathrm{(E) \ } 194 </math>
-== Solution ==
-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\mathrm{ (B) \ }</math>