Problem:
It is given that log6​a+log6​b+log6​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.
Solution:
Note that 6=log6​a+log6​b+log6​c=log6​abc. Then 66=abc=b3, so b=62 and ac=64. Because bî€ =a and b−a 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=111​.
The problems on this page are the property of the MAA's American Mathematics Competitions