AoPSWiki
Want to learn how to tackle those tough AMC/AIME/Olympiad algebra problems? Check out Art of Problem Solving's Intermediate Algebra by Richard Rusczyk and Mathew Crawford. Over 1600 problems!

2002 AIME II Problems/Problem 3

From AoPSWiki

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}

See also

2002 AIME II (ProblemsResources)
Preceded by
Problem 2
Followed by
Problem 4
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
Trying to get to the USAMO in 2010? Our AIME Problem Series can help you get there! Click here to enroll today!
© Copyright 2008 AoPS Incorporated. All Rights Reserved. • FoundationPrivacyContact Us