Solution #34

We have log₆a + log₆b + log₆c = 6, so log₆abc = 6.

Thus abc = 6⁶. Because a, b, c is an geometric sequence, we have b = ra, and c = r²a for some r > 1.

Thus abc = r³a³, so r³a³ = 6⁶, and ra = 36.

ra = b, so b is 36. The difference between b and a is a perfect square, so the possible values for a are 35, 32, 27, 20, and 11.

Each value for a generates a value for r, and we can use that r value to see if it produces an integer value for c. The only value that works is 27, which makes r = 4/3, and makes c = 48. Thus a + b + c = 27 + 36 + 48 = 111.