Difference between revisions of "2002 AMC 12B Problems/Problem 7"
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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^ | + | 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- | + | {{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 times their sum. What is the sum of their squares?
Solution
Let the three consecutive integers be ; then
Since , we have , with the positive solution being . Then .
See also
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 |