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 .
Like above, we use logarithmic identities to convert the problem into workable equations. We begin by labelling the powers of 3. Call , , ..., as , , and ... respectively. With this format we can rewrite the first given equation as . Simplify to get . (1)
Now, rewrite the second given equation as . Obviously, , aka because there are some small fractional change that is left over. This means is . Thinking about the geometric sequence, it's clear each consecutive value of will be either a power of three times smaller or larger. In other words, the earliest values of will be negligible compared to the last values of . Even in the best case scenario, where the common ratio is 3, the values left of are not enough to sum to a value greater than 2 times (amount needed to raise the power of 3 by 1). This confirms that . (2)
Use equations 1 and 2 to get and .
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