2002 AIME II Problem 3

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

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.

Problem 3 in other years