Difference between revisions of "2002 AMC 12B Problems/Problem 7"
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\qquad\mathrm{(E)}\ 194</math> | \qquad\mathrm{(E)}\ 194</math> | ||
== 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 | + | 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)=24a\Rightarrow(a-1)(a+1)=24</math>. <math>24=4\times6</math>, so <math>a=5</math>. Hence, the sum of the squares is <math>4^2+5^2+6^2=\boxed{\mathrm{ (B)}\ 77}</math>. |
== See also == | == See also == | ||
Revision as of 23:48, 15 July 2014
- The following problem is from both the 2002 AMC 12B #7 and 2002 AMC 10B #11, so both problems redirect to this page.
Problem
The product of three consecutive positive integers is 
times their sum. What is the sum of their squares?
Solution
Let the three consecutive positive integers be 
, 
, and 
. So, 
. 
, 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 | ||
| 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 | |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
