Difference between revisions of "2002 AIME II Problems/Problem 3"
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<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>. | ||
| − | Since <math>b-a</math> is the square of an integer, we can find a few values of <math>a</math> that work: 11, 20, 27, 32, and 35. 11 doesn't work. Nor do 20, 32, or 35. Thus, <math>a=27</math>, and <math>c=\dfrac{36}{27}\cdot 36=\dfrac{4}{3}\cdot 36 | + | Since <math>b-a</math> is the square of an integer, we can find a few values of <math>a</math> that work: 11, 20, 27, 32, and 35. 11 doesn't work. Nor do 20, 32, or 35. Thus, <math>a=27</math>, and <math>c=\dfrac{36}{27}\cdot 36=\dfrac{4}{3}\cdot 36=48</math>. |
<math>a+b+c=27+36+48=\boxed{111}</math> | <math>a+b+c=27+36+48=\boxed{111}</math> | ||
Revision as of 13:24, 12 March 2015
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 
.
See also
| 2002 AIME II (Problems • Answer Key • Resources) | ||
| Preceded by Problem 2 |
Followed by Problem 4 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
| All AIME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
