2002 AIME II 第 3 题

Combine the logs: $abc = 6^6.$ In a geometric sequence $ac = b^2,$ so $b^3 = 6^6$

With $b = 36,$ write $a = 36 - k^2$ and require that $a$ divide $36^2$

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 $c = \frac{1296}{27} = 48.$

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