Difference between revisions of "2002 AIME II Problems/Problem 3"
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== Problem == | == Problem == | ||
| − | It is given that <math>\log_{6}a + \log_{6}b + \log_{6}c = 6,</math> where <math>a,</math> <math>b,</math> and <math>c</math> are [[positive]] [[integer]]s that form an increasing [[geometric sequence]] and <math>b - a</math> is the [[square]] of an integer. Find <math>a + b + c.</math> | + | It is given that <math>\log_{6}a + \log_{6}b + \log_{6}c = 6,</math> where <math>a,</math> <math>b,</math> and <math>c</math> are [[positive]] [[integer]]s that form an increasing [[geometric sequence]] and <math>b - a</math> is the [[Perfect square|square]] of an integer. Find <math>a + b + c.</math> |
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== Solution == | == Solution == | ||
<math>abc=6^6</math>. Since they form an increasing geometric sequence, <math>b</math> is the [[geometric mean]] of the [[product]] <math>abc</math>. <math>b=\sqrt[3]{abc}=6^2=36</math>. | <math>abc=6^6</math>. Since they form an increasing geometric sequence, <math>b</math> is the [[geometric mean]] of the [[product]] <math>abc</math>. <math>b=\sqrt[3]{abc}=6^2=36</math>. | ||
Revision as of 07:43, 17 April 2008
Problem
It is given that 
where 
and 
are positive integers that form an increasing geometric sequence and 
is the square of an integer. Find 
Solution
. Since they form an increasing geometric sequence, 
is the geometric mean of the product 
. ![$b=\sqrt[3]{abc}=6^2=36$](http://latex.artofproblemsolving.com/8/7/f/87ffcb66a32127aa794a8e3f2db141b6fe092d09.png)
.
Since 
is the square of an integer, we can find a few values of 
that work: 11, 20, 27, 32, and 35. 11 doesn't work. Nor do 20, 32, or 35. Thus, 
, and $c=\dfrac{36}{27}\cdot 36=\dfrac{4}{3}\cdot 36}=48$ (Error compiling LaTeX. Unknown error_msg)
![\[a+b+c=27+36+48=\boxed{111}\]](http://latex.artofproblemsolving.com/7/b/c/7bcb1f92b5f3779759bbc39fd1ff45b3690c36a7.png)
