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>. | ||
+ | |||
+ | ==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 14:52, 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.