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>. | ||
+ | |||
+ | ==Solution 3== | ||
+ | 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 \left( \sum_{n=0}^{7}x_{n} \right) \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> | ||
+ | |||
+ | |||
+ | -jackshi2006 | ||
+ | ==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 == | ||
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[[Category:Intermediate Algebra Problems]] | [[Category:Intermediate Algebra Problems]] | ||
+ | {{MAA Notice}} |
Latest revision as of 11:22, 8 January 2024
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
Solution 4 (dum)
Proceed as in Solution 3 for the first few steps. We have the sequence . As stated above, we then get that
, from which we simplify to
. From here, we just go brute force using the second statement (that
). Rearranging the equation from earlier, we get
from which it is clear that
is a multiple of
. Testing the first few values of
, we get:
Case 1 (
)
The sequence is then
, which breaks the upper bound.
Case 2 (
)
The sequence is then
, which also breaks the upper bound.
Case 3 (
)
This is the first reasonable one, giving
. It seems like this would break the upper bound, but from some testing we get:
Repeating this process over and over, we eventually get (while ignoring the extremely small fractions left over)
which confirms that this satisfies our upper bound. Thus
, so
. We then get the requested answer,
~ 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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