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Difference between revisions of "2002 AIME II Problems/Problem 3"
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== Problem ==
 
== Problem ==
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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 integers that form an increasing geometric sequence and <math>b - a</math> is the square of an integer. Find <math>a + b + c.</math>
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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 ==
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{{solution}}
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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>.
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Since <math>b-a</math> is the square of an integer, we can find a few values of <math>a</math> that work: <math>11, 20, 27, 32,</math> and <math>35</math>. Out of these, the only value of <math>a</math> that works is <math>a=27</math>, from which we can deduce that <math>c=\dfrac{36}{27}\cdot 36=\dfrac{4}{3}\cdot 36=48</math>.
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Thus, <math>a+b+c=27+36+48=\boxed{111}</math>
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== See also ==
 
== See also ==
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* [[2002 AIME II Problems/Problem 2 | Previous problem]]
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{{AIME box|year=2002|n=II|num-b=2|num-a=4}}
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* [[2002 AIME II Problems/Problem 4 | Next problem]]
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* [[2002 AIME II Problems]]
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[[Category: Intermediate Algebra Problems]]
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{{MAA Notice}}

Revision as of 03:31, 6 December 2019

Problem

It is given that $\log_{6}a + \log_{6}b + \log_{6}c = 6,$ where $a,$ $b,$ and $c$ are positive integers that form an increasing geometric sequence and $b - a$ is the square of an integer. Find $a + b + c.$

Solution

$abc=6^6$. Since they form an increasing geometric sequence, $b$ is the geometric mean of the product $abc$. $b=\sqrt[3]{abc}=6^2=36$.

Since $b-a$ is the square of an integer, we can find a few values of $a$ that work: $11, 20, 27, 32,$ and $35$. Out of these, the only value of $a$ that works is $a=27$, from which we can deduce that $c=\dfrac{36}{27}\cdot 36=\dfrac{4}{3}\cdot 36=48$.

Thus, $a+b+c=27+36+48=\boxed{111}$

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

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