2002 AIME II Problems/Problem 3

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. 11 doesn't work. Nor do 20, 32, or 35. Thus, $a=27$ , and $c=\dfrac{36}{27}\cdot 36=\dfrac{4}{3}\cdot 36=48$ .

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

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
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2002 AIME II Problems/Problem 3

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