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 $c=\dfrac{36}{27}\cdot 36=\dfrac{4}{3}\cdot 36}=48$ (Error compiling LaTeX. Unknown error_msg)

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

Page

Toolbox

Search

2002 AIME II Problems/Problem 3

Problem

Solution

See also