Difference between revisions of "2006 AIME I Problems/Problem 9"
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<math>y=\frac{1003-2x}{11}</math> | <math>y=\frac{1003-2x}{11}</math> | ||
− | For y to be an integer, the numerator must be divisible by 11. This occurs when <math>x=1</math> because <math>1001=91*11</math>. Because only even | + | For <math>y</math> to be an integer, the [[numerator]] must be [[divisible]] by 11. This occurs when <math>x=1</math> because <math>1001=91*11</math>. Because only [[even integer]]s are being subtracted from 1003, the numerator never equals an even [[multiple]] of 11. Therefore, the numerator takes on the value of every [[odd integer | odd]] multiple of 11 from 11 to 1001. Since the odd multiples are separated by a distance of 22, the number of ordered pairs that work is <math>1 + \frac{1001-11}{22}=1 + \frac{990}{22}=46</math>. (We must add 1 because both endpoints are being included.) So the answer is <math>46</math>. |
== See also == | == See also == |
Revision as of 16:51, 12 October 2006
Problem
The sequence is geometric with
and common ratio
where
and
are positive integers. Given that
find the number of possible ordered pairs
Solution
So our question is equivalent to solving
![$\log_8 (a^{12}r^{66})=2006$](http://latex.artofproblemsolving.com/1/c/8/1c8d6cbd59decadfd232d4fab79e93e2def4115d.png)
for positive integers.
so
The product of and
is a power of 2. Since both numbers have to be integers, this means that
and
are themselves powers of 2. Now, let
and
:
For to be an integer, the numerator must be divisible by 11. This occurs when
because
. Because only even integers are being subtracted from 1003, the numerator never equals an even multiple of 11. Therefore, the numerator takes on the value of every odd multiple of 11 from 11 to 1001. Since the odd multiples are separated by a distance of 22, the number of ordered pairs that work is
. (We must add 1 because both endpoints are being included.) So the answer is
.