Difference between revisions of "2007 AIME II Problems/Problem 12"
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<math>\sum_{n=0}^{7}\log_{3}(x_{n}) = 308</math> and <math>56 \leq \log_{3}\left ( \sum_{n=0}^{7}x_{n}\right ) \leq 57,</math> | <math>\sum_{n=0}^{7}\log_{3}(x_{n}) = 308</math> and <math>56 \leq \log_{3}\left ( \sum_{n=0}^{7}x_{n}\right ) \leq 57,</math> | ||
− | find <math> | + | find <math>\log_{3}(x_{14}).</math> |
== Solution == | == Solution == | ||
− | Suppose that <math> | + | Suppose that <math>x_0 = a</math>, and that the common [[ratio]] between the terms is <math>r</math>. |
− | The first conditions tells us that <math> | + | The first conditions tells us that <math>\log_3 a + \log_3 ar + \ldots + \log_3 ar^7 = 308</math>. Using the rules of [[logarithm]]s, we can simplify that to <math>\log_3 a^8r^{1 + 2 + \ldots + 7} = 308</math>. Thus, <math>a^8r^{28} = 3^{308}</math>. Since all of the terms of the geometric sequence are integral powers of <math>3</math>, we know that both <math>a</math> and <math>r</math> must be powers of 3. Denote <math>3^x = a</math> and <math>3^y = r</math>. We find that <math>8x + 28y = 308</math>. The possible positive integral pairs of <math>(x,y)</math> are <math>(35,1),\ (28,3),\ (21,5),\ (14,7),\ (7,9),\ (0,11)</math>. |
− | The second condition tells us that <math>56 \le \log_3 (a + ar + \ldots ar^7) \le 57</math>. Using the sum formula for a [[geometric series]] and substituting <math>x</math> and <math>y</math>, this simplifies to <math>3^{56} \le 3^x \frac{3^{8y} - 1}{3^y-1} \le 3^{57}</math>. The fractional part <math>\approx \frac{3^{8y}}{3^y} = 3^{7y}</math>. Thus, we need <math>\approx 56 \le x + 7y \le 57</math>. Checking the pairs above, only <math> | + | The second condition tells us that <math>56 \le \log_3 (a + ar + \ldots ar^7) \le 57</math>. Using the sum formula for a [[geometric series]] and substituting <math>x</math> and <math>y</math>, this simplifies to <math>3^{56} \le 3^x \frac{3^{8y} - 1}{3^y-1} \le 3^{57}</math>. The fractional part <math>\approx \frac{3^{8y}}{3^y} = 3^{7y}</math>. Thus, we need <math>\approx 56 \le x + 7y \le 57</math>. Checking the pairs above, only <math>(21,5)</math> is close. |
− | Our solution is therefore <math>\log_3 (ar^{14}) = \log_3 3^x + \log_3 3^{14y} = x + 14y = 091 \ | + | Our solution is therefore <math>\log_3 (ar^{14}) = \log_3 3^x + \log_3 3^{14y} = x + 14y = \boxed{091}</math>. |
+ | |||
+ | ==Solution 2== | ||
+ | All these integral powers of <math>3</math> are all different, thus in base <math>3</math> the sum of these powers would consist of <math>1</math>s and <math>0</math>s. Thus the largest value <math>x_7</math> must be <math>3^{56}</math> in order to preserve the givens. Then we find by the given that <math>x_7x_6x_5\dots x_0 = 3^{308}</math>, and we know that the exponents of <math>x_i</math> are in an arithmetic sequence. Thus <math>56+(56-r)+(56-2r)+\dots +(56-7r) = 308</math>, and <math>r = 5</math>. Thus <math>\log_3 (x_{14}) = \boxed{091}</math>. | ||
== See also == | == See also == | ||
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[[Category:Intermediate Algebra Problems]] | [[Category:Intermediate Algebra Problems]] | ||
+ | {{MAA Notice}} |
Revision as of 14:53, 20 February 2017
Contents
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 .
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.