Difference between revisions of "2007 AIME II Problems/Problem 12"
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Our solution is therefore <math>\log_3 (ar^{14}) = \log_3 3^x + \log_3 3^{14y} = x + 14y = \boxed{091}</math>. | Our solution is therefore <math>\log_3 (ar^{14}) = \log_3 3^x + \log_3 3^{14y} = x + 14y = \boxed{091}</math>. | ||
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| + | ==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 == | ||
Revision as of 13:52, 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. 
