Difference between revisions of "2007 AIME II Problems/Problem 11"
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| − | Our solution is therefore <math> | + | Our solution is therefore <math>\log_3 (ar^{14}) = \log_3 3^x + \log_3 3^{14y} = x + 14y = 091 \displaystyle</math>. |
== See also == | == See also == | ||
Revision as of 16:17, 30 March 2007
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 
.
See also
| 2007 AIME II (Problems • Answer Key • Resources) | ||
| Preceded by Problem 10 |
Followed by Problem 12 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
| All AIME Problems and Solutions | ||
