Difference between revisions of "2007 AIME II Problems/Problem 12"
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Like above, we use logarithmic identities to convert the problem into workable equations. We begin by labelling the powers of 3. Call <math>x_0</math>, <math>x_1</math>, <math>x_2</math>..., as <math>3^n</math>, <math>3^{n+m}</math>, and <math>3^{n+2m}</math>... respectively. With this format we can rewrite the first given equation as <math>n + n + m + n+2m + n+3m+...+n+7m = 308</math>. Simplify to get <math>2n+7m=77</math>. (1) | Like above, we use logarithmic identities to convert the problem into workable equations. We begin by labelling the powers of 3. Call <math>x_0</math>, <math>x_1</math>, <math>x_2</math>..., as <math>3^n</math>, <math>3^{n+m}</math>, and <math>3^{n+2m}</math>... respectively. With this format we can rewrite the first given equation as <math>n + n + m + n+2m + n+3m+...+n+7m = 308</math>. Simplify to get <math>2n+7m=77</math>. (1) | ||
− | Now, rewrite the second given equation as <math>3^{56} \leq ( \sum_{n=0}^{7}x_{n} ) \leq 3^{57}</math>. Obviously, <math>x_7</math>, aka <math>3^{n+7m}</math> <math><3^{57}</math> because there are some small fractional change that is left over. This means <math>n+7m</math> is <math>\leq56</math>. Thinking about the geometric sequence, it's clear each consecutive value of <math>x_i</math> will be either a power of three times smaller or larger. | + | Now, rewrite the second given equation as <math>3^{56} \leq ( \sum_{n=0}^{7}x_{n} ) \leq 3^{57}</math>. Obviously, <math>x_7</math>, aka <math>3^{n+7m}</math> <math><3^{57}</math> because there are some small fractional change that is left over. This means <math>n+7m</math> is <math>\leq56</math>. Thinking about the geometric sequence, it's clear each consecutive value of <math>x_i</math> will be either a power of three times smaller or larger. In other words, the earliest values of <math>x_i</math> will be negligible compared to the last values of <math>x_i</math>. Even in the best case scenario, where the common ratio is 3), the values left of <math>x_7</math> are not enough to sum to a value greater than 2 times <math>x_7</math> (amount needed to raise the power of 3 by 1. This confirms that <math>3^{n+7m} = 3^{56}</math>. (2) |
Use equations 1 and 2 to get <math>m=5</math> and <math>n=21</math>. <math>log_{3}{x_{14}} = log_{3}{3^{21+14*5}} = 21+14*5 = \boxed{091}</math> | Use equations 1 and 2 to get <math>m=5</math> and <math>n=21</math>. <math>log_{3}{x_{14}} = log_{3}{3^{21+14*5}} = 21+14*5 = \boxed{091}</math> | ||
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+ | -jackshi2006 | ||
== See also == | == See also == |
Revision as of 19:25, 29 December 2020
Problem
The increasing geometric sequence consists entirely of integral powers of Given that
and
find
Solution
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 .
Solution 2
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 .
Solution 3
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 .
-jackshi2006
See also
2007 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 11 |
Followed by Problem 13 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.