Question

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 $a + b + c.$

Answer

Answer: 111

Concepts:logarithm geometric sequence divisibility

Difficulty rating: 2170

Solution:

Adding the logs gives $\log_6(abc) = 6,$ so $abc = 6^6.$ In a geometric sequence $ac = b^2,$ hence $b^3 = 6^6,$ so $b = 36$ and $ac = 36^2 = 1296.$

Since the sequence is increasing, $b - a$ is a positive perfect square, so $a = 36 - k^2$ for some $k = 1, \ldots, 5,$ giving candidates $35, 32, 27, 20, 11.$ Also $a$ must divide $1296 = 2^4 \cdot 3^4,$ and of the candidates only $27$ does, with $c = 1296/27 = 48.$

Indeed $27, 36, 48$ is geometric with ratio $\frac{4}{3},$ and $a + b + c = 27 + 36 + 48 = 111.$

2002 AIME II Problem 3

Answer: 111

Difficulty rating: 2170

Solution:

Problem 3 in Other Years