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Difference between revisions of "2007 AIME II Problems/Problem 11"

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== Solution ==
 
== Solution ==
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Suppose that <math>x_0 = a</math>, and that the common [[ratio]] between the terms is <math>r</math>.  
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Suppose that <math>\displaystyle x_0 = a</math>, and that the common [[ratio]] between the terms is <math>r</math>.  
  
  
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The first conditions tells us that <math>\displaystyle \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>\displaystyle \log_3 a^8r^{1 \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 integral pairs of <math>(x,y)</math> are <math>(35,1),\ (28,3),\ (21,5),\ (14,7),\ (7,9),\ (0,11)</math>.
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The first conditions tells us that <math>\displaystyle \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>\displaystyle \log_3 a^8r^{1 + 2 + \ldots + 7} = 308</math>. Thus, <math>\displaystyle 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>\displaystyle 3^x = a</math> and <math>\displaystyle 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>.
  
  
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Our solution is therefore <math>\displaystyle \log_3 (ar^{14}) = \log_3 3^x + \log_3 3^{14y} = x + 14y = 91</math>.
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Our solution is therefore <math>\displaystyle \log_3 (ar^{14}) = \log_3 3^x + \log_3 3^{14y} = x + 14y = 091</math>.
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== See also ==
 
== See also ==
 
{{AIME box|year=2007|n=II|num-b=10|num-a=12}}
 
{{AIME box|year=2007|n=II|num-b=10|num-a=12}}
  
 
[[Category:Intermediate Algebra Problems]]
 
[[Category:Intermediate Algebra Problems]]

Revision as of 16:17, 30 March 2007

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 $\displaystyle \log_3 (ar^{14}) = \log_3 3^x + \log_3 3^{14y} = x + 14y = 091$.

See also

2007 AIME II (Problems • Answer Key • Resources)
Preceded by
Problem 10 		Followed by
Problem 12
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions
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