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

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m (Reverted edits by Mathheads1 (talk) to last revision by Azjps)
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<cmath>(x-1)(x)(x+1) = x(x^2 - 1) = 8(x-1 + x + x+1) = 24x</cmath>
 
<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 ==
 
== See also ==
{{AMC12 box|year=2002|ab=B|num-b=6|num-876897689798=8}}
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{{AMC12 box|year=2002|ab=B|num-b=6|num-a=8}}
  
 
[[Category:Introductory Algebra Problems]]
 
[[Category:Introductory Algebra Problems]]

Revision as of 22:46, 6 June 2011

Problem

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

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

Solution

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 $4^2 + 5^2 + 6^2 = 77\ \mathrm{(B)}$.

See also

2002 AMC 12B (ProblemsAnswer KeyResources)
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
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