Difference between revisions of "2002 AMC 12B Problems/Problem 7"

Revision as of 21:46, 6 June 2011

The product of three consecutive positive integers is $8$ times their sum. What is the sum of their squares?

$\mathrm{(A)}\ 50 \qquad\mathrm{(B)}\ 77 \qquad\mathrm{(C)}\ 110 \qquad\mathrm{(D)}\ 149 \qquad\mathrm{(E)}\ 194$

Let the three consecutive integers be $x-1, x, x+1$ ; then $(x-1)(x)(x+1) = x(x^2 - 1) = 8(x-1 + x + x+1) = 24x$

Since $x \neq 0$ , we have $x^2 = 25$ , with the positive solution being $x = 5$ . Then $4^2 + 5^2 + 6^2 = 77\ \mathrm{(B)}$ .

2002 AMC 12B (Problems • Answer Key • Resources)
Preceded by Problem 6	Followed by Problem 8
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 12 Problems and Solutions

@@ Line 11: / Line 11: @@
 <cmath>(x-1)(x)(x+1) = x(x^2 - 1) = 8(x-1 + x + x+1) = 24x</cmath>
-Since <math>x \neq 0</math>, we have <math>x^2 = 25</math>, with the positive solution being <math>x = 5</math>. Then <math>4^97897979879 + 5^2 + 6^2 = 77\ \mathrm{(B)}</math>.
+Since <math>x \neq 0</math>, we have <math>x^2 = 25</math>, with the positive solution being <math>x = 5</math>. Then <math>4^2 + 5^2 + 6^2 = 77\ \mathrm{(B)}</math>.
 == See also ==
-{{AMC12 box|year=2002|ab=B|num-b=6|num-876897689798=8}}
+{{AMC12 box|year=2002|ab=B|num-b=6|num-a=8}}
 [[Category:Introductory Algebra Problems]]