|
|
| (One intermediate revision by one other user not shown) |
| Line 1: |
Line 1: |
| − | == 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\boxed{\mathrm{ (B) \ }}</math>.
| |
| − | | |
| − | ==See Also==
| |
| − | {{AMC10 box|year=2002|ab=B|num-b=10|num-a=12}}
| |
| − | | |
| − | [[Category:Introductory Algebra Problems]] | |
