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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]] [[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>
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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 ==
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== Solution 1==
 
<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>.
  
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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: 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>
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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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==Solution 2(similar to Solution 1)==
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Let <math>r</math> be the common ratio of the geometric sequence. Since it is increasing, that means that <math>b = ar</math>, and <math>c = ar^2</math>. Simplifying the logarithm, we get <math>\log_6(a^3*r^3) = 6</math>. Therefore, <math>a^3*r^3 = 6^6</math>. Taking the cube root of both sides, we see that <math>ar = 6^2 = 36</math>. Now since <math>ar = b</math>, that means <math>b = 36</math>. Using the trial and error shown in solution 1, we get <math>a = 27</math>, and <math>r = \frac{4}{3}</math>. Now, <math>27*r^2= c = 48</math>. Therefore, the answer is <math>27+36+48 = \boxed{111}</math>
  
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<cmath>a+b+c=27+36+48=\boxed{111}</cmath>
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~idk12345678
  
 
== 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}}

Latest revision as of 21:50, 5 April 2024

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 1

$abc=6^6$. 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, $a+b+c=27+36+48=\boxed{111}$

Solution 2(similar to Solution 1)

Let $r$ be the common ratio of the geometric sequence. Since it is increasing, that means that $b = ar$, and $c = ar^2$. Simplifying the logarithm, we get $\log_6(a^3*r^3) = 6$. Therefore, $a^3*r^3 = 6^6$. Taking the cube root of both sides, we see that $ar = 6^2 = 36$. Now since $ar = b$, that means $b = 36$. Using the trial and error shown in solution 1, we get $a = 27$, and $r = \frac{4}{3}$. Now, $27*r^2= c = 48$. Therefore, the answer is $27+36+48 = \boxed{111}$

~idk12345678

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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