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2007 AIME II Problems/Problem 12

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Problem

The increasing geometric sequence x_{0},x_{1},x_{2},\ldots consists entirely of integral powers of 3. Given that

\sum_{n=0}^{7}\log_{3}(x_{n}) = 308 and 56 \leq \log_{3}\left ( \sum_{n=0}^{7}x_{n}\right ) \leq 57,

find \displaystyle \log_{3}(x_{14}).

Solution

Suppose that \displaystyle x_0 = a, and that the common ratio between the terms is r.


The first conditions tells us that \displaystyle \log_3 a + \log_3 ar \ldots \log_3 ar^7 = 308. Using the rules of logarithms, we can simplify that to \displaystyle \log_3 a^8r^{1 + 2 + \ldots + 7} = 308. Thus, \displaystyle a^8r^{28} = 3^{308}. Since all of the terms of the geometric sequence are integral powers of 3, we know that both a and r must be powers of 3. Denote \displaystyle 3^x = a and \displaystyle 3^y = r. We find that 8x + 28y = 308. The possible positive integral pairs of (x,y) are (35,1),\ (28,3),\ (21,5),\ (14,7),\ (7,9),\ (0,11).


The second condition tells us that 56 \le \log_3 (a + ar + \ldots ar^7) \le 57. Using the sum formula for a geometric series and substituting x and y, this simplifies to 3^{56} \le 3^x \frac{3^{8y} - 1}{3^y-1} \le 3^{57}. The fractional part \approx \frac{3^{8y}}{3^y} = 3^{7y}. Thus, we need \approx 56 \le x + 7y \le 57. Checking the pairs above, only \displaystyle (21,5) is close.


Our solution is therefore \log_3 (ar^{14}) = \log_3 3^x + \log_3 3^{14y} = x + 14y = 091 \displaystyle.

See also

2007 AIME II (ProblemsResources)
Preceded by
Problem 11
Followed by
Problem 13
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
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