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Difference between revisions of "2007 AIME II Problems/Problem 12"

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==Solution 2==
 
==Solution 2==
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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>.
+
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 ==

Revision as of 13:53, 20 February 2017

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 $\log_{3}(x_{14}).$

Solution

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


The first conditions tells us that $\log_3 a + \log_3 ar + \ldots + \log_3 ar^7 = 308$. Using the rules of logarithms, we can simplify that to $\log_3 a^8r^{1 + 2 + \ldots + 7} = 308$. Thus, $a^8r^{28} = 3^{308}$. 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 $3^x = a$ and $3^y = r$. 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 $(21,5)$ is close.


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

Solution 2

All these integral powers of $3$ are all different, thus in base $3$ the sum of these powers would consist of $1$s and $0$s. Thus the largest value $x_7$ must be $3^{56}$ in order to preserve the givens. Then we find by the given that $x_7x_6x_5\dots x_0 = 3^{308}$, and we know that the exponents of $x_i$ are in an arithmetic sequence. Thus $56+(56-r)+(56-2r)+\dots +(56-7r) = 308$, and $r = 5$. Thus $\log_3 (x_{14}) = \boxed{091}$.

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

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