Difference between revisions of "2002 AMC 10B Problems/Problem 11"
(New page: == Problem == 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\qqu...) |
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== Solution == | == 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> | + | 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>. |
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| + | ==See Also== | ||
| + | {{AMC10 box|year=2002|ab=B|num-b=10|num-a=12}} | ||
| + | |||
| + | [[Category:Introductory Algebra Problems]] | ||
Revision as of 14:12, 27 December 2008
Problem
The product of three consecutive positive integers is 
times their sum. What is the sum of the squares?
Solution
Let the three consecutive positive integers be 
, 
, and 
. So, 
. Rearranging and factoring, 
, so 
. Hence, the sum of the squares is 
.
See Also
| 2002 AMC 10B (Problems • Answer Key • Resources) | ||
| Preceded by Problem 10 |
Followed by Problem 12 | |
| 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 10 Problems and Solutions | ||
