2007 AIME II Problems/Problem 12
Suppose that , and that the common ratio between the terms is .
The first conditions tells us that . Using the rules of logarithms, we can simplify that to . Thus, . Since all of the terms of the geometric sequence are integral powers of , we know that both and must be powers of 3. Denote and . We find that . The possible positive integral pairs of are .
The second condition tells us that . Using the sum formula for a geometric series and substituting and , this simplifies to . The fractional part . Thus, we need . Checking the pairs above, only is close.
Our solution is therefore .
All these integral powers of are all different, thus in base the sum of these powers would consist of s and s. Thus the largest value must be in order to preserve the givens. Then we find by the given that , and we know that the exponents of are in an arithmetic sequence. Thus , and . Thus .
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