Problem:
It is given that log6βa+log6βb+log6βc=6, where a,b, and c are positive integers that form an increasing geometric sequence and bβa is the square of an integer. Find a+b+c.
Solution:
Note that 6=log6βa+log6βb+log6βc=log6βabc. Then 66=abc=b3, so b=62 and ac=64. Because bξ =a and bβa is the square of an integer, the only possibilities for a are 11,20,27,32, and 35. Of these, only 27 is a divisor of 64. Thus a+b+c=27+36+48=111β.
The problems on this page are the property of the MAA's American Mathematics Competitions