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

(→‎Solution 3)
(→‎Solution 3)
 
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-jackshi2006
 
-jackshi2006
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==Solution 4 (dum)==
 +
Proceed as in Solution 3 for the first few steps. We have the sequence <math>3^{a},3^{a+n},3^{a+2n}...</math>. As stated above, we then get that <math>a+a+n+...+a+7n=308</math>, from which we simplify to <math>2a+7n=77</math>. From here, we just go brute force using the second statement (that <math>3^{56}\leq 3^{a}+...+3^{a+7n}\leq 3^{57}</math>). Rearranging the equation from earlier, we get
 +
<cmath>n=11-\frac{2a}{7}</cmath>
 +
from which it is clear that <math>a</math> is a multiple of <math>7</math>. Testing the first few values of <math>a</math>, we get:
 +
Case 1 (<math>a=7,n=9</math>)
 +
The sequence is then <math>3^{7}+...+3^{70}</math>, which breaks the upper bound.
 +
Case 2 (<math>a=14,n=7</math>)
 +
The sequence is then <math>3^{14}+...+3^{63}</math>, which also breaks the upper bound.
 +
Case 3 (<math>a=21,n=5</math>)
 +
This is the first reasonable one, giving <math>3^{21}+...+3^{56}</math>. It seems like this would break the upper bound, but from some testing we get:
 +
<cmath>3^{21}+...+3^{56}<3^{57}</cmath>
 +
<cmath>1+...+3^{35}<3^{36}</cmath>
 +
<cmath>1+...+3^{30}<2*3^{35}</cmath>
 +
<cmath>3^{5}+...+3^{30}<2*3^{35}-1</cmath>
 +
<cmath>1+...+3^{25}<2*3^{30}-3^{-5}</cmath>
 +
Repeating this process over and over, we eventually get (while ignoring the extremely small fractions left over)
 +
<cmath>1<2*3^{5}</cmath>
 +
which confirms that this satisfies our upper bound. Thus <math>a=21,n=5</math>, so <math>x_14=3^{a+14n}\rightarrow3^{91}</math>. We then get the requested answer, <math>\log_3(3^{91})=\boxed{091}</math> ~ amcrunner
  
 
== See also ==
 
== See also ==

Latest revision as of 11:22, 8 January 2024

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 $\log_3 (x_{14}) = \boxed{091}$.

Solution 3

Like above, we use logarithmic identities to convert the problem into workable equations. We begin by labelling the powers of 3. Call $x_0$, $x_1$, $x_2$..., as $3^n$, $3^{n+m}$, and $3^{n+2m}$... respectively. With this format we can rewrite the first given equation as $n + n + m + n+2m + n+3m+...+n+7m = 308$. Simplify to get $2n+7m=77$. (1)

Now, rewrite the second given equation as $3^{56} \leq \left( \sum_{n=0}^{7}x_{n} \right) \leq 3^{57}$. Obviously, $x_7$, aka $3^{n+7m}$ $<3^{57}$ because there are some small fractional change that is left over. This means $n+7m$ is $\leq56$. Thinking about the geometric sequence, it's clear each consecutive value of $x_i$ will be either a power of three times smaller or larger. In other words, the earliest values of $x_i$ will be negligible compared to the last values of $x_i$. Even in the best case scenario, where the common ratio is 3, the values left of $x_7$ are not enough to sum to a value greater than 2 times $x_7$ (amount needed to raise the power of 3 by 1). This confirms that $3^{n+7m} = 3^{56}$. (2)

Use equations 1 and 2 to get $m=5$ and $n=21$. $\log_{3}{x_{14}} = \log_{3}{3^{21+14*5}} = 21+14*5 = \boxed{091}$


-jackshi2006

Solution 4 (dum)

Proceed as in Solution 3 for the first few steps. We have the sequence $3^{a},3^{a+n},3^{a+2n}...$. As stated above, we then get that $a+a+n+...+a+7n=308$, from which we simplify to $2a+7n=77$. From here, we just go brute force using the second statement (that $3^{56}\leq 3^{a}+...+3^{a+7n}\leq 3^{57}$). Rearranging the equation from earlier, we get \[n=11-\frac{2a}{7}\] from which it is clear that $a$ is a multiple of $7$. Testing the first few values of $a$, we get: Case 1 ($a=7,n=9$) The sequence is then $3^{7}+...+3^{70}$, which breaks the upper bound. Case 2 ($a=14,n=7$) The sequence is then $3^{14}+...+3^{63}$, which also breaks the upper bound. Case 3 ($a=21,n=5$) This is the first reasonable one, giving $3^{21}+...+3^{56}$. It seems like this would break the upper bound, but from some testing we get: \[3^{21}+...+3^{56}<3^{57}\] \[1+...+3^{35}<3^{36}\] \[1+...+3^{30}<2*3^{35}\] \[3^{5}+...+3^{30}<2*3^{35}-1\] \[1+...+3^{25}<2*3^{30}-3^{-5}\] Repeating this process over and over, we eventually get (while ignoring the extremely small fractions left over) \[1<2*3^{5}\] which confirms that this satisfies our upper bound. Thus $a=21,n=5$, so $x_14=3^{a+14n}\rightarrow3^{91}$. We then get the requested answer, $\log_3(3^{91})=\boxed{091}$ ~ amcrunner

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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