Difference between revisions of "2006 AIME I Problems/Problem 9"
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− | For the step above, you may also simply do <math>1001/11 + 1 = 91 + 1 = 92</math> to find how many multiples of <math>11</math> there are in between <math>11</math> and <math>1001</math>. Then, divide <math>92/2</math> to find only the odd solutions. <math>-XxHalo711</math> | + | For the step above, you may also simply do <math>1001/11 + 1 = 91 + 1 = 92</math> to find how many multiples of <math>11</math> there are in between <math>11</math> and <math>1001</math>. Then, divide <math>92/2</math> = <math>\boxed{046}</math> to find only the odd solutions. <math>-XxHalo711</math> |
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Revision as of 23:33, 15 June 2019
Contents
Problem
The sequence is geometric with
and common ratio
where
and
are positive integers. Given that
find the number of possible ordered pairs
Solution 1
So our question is equivalent to solving 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
. This occurs when
because
. Because only even integers are being subtracted from
, the numerator never equals an even multiple of
. Therefore, the numerator takes on the value of every odd multiple of
from
to
. Since the odd multiples are separated by a distance of
, the number of ordered pairs that work is
. (We must add 1 because both endpoints are being included.) So the answer is
.
For the step above, you may also simply do to find how many multiples of
there are in between
and
. Then, divide
=
to find only the odd solutions.
Another way is to write
Since , the answer is just the number of odd integers in
, which is, again,
.
Solution 2
Using the above method, we can derive that .
Now, think about what happens when r is an even power of 2. Then
must be an odd power of 2 in order to satisfy the equation which is clearly not possible. Thus the only restriction r has is that it must be an odd power of 2, so
,
,
.... all work for r, until r hits
, when it gets greater than
, so the greatest value for r is
. All that's left is to count the number of odd integers between 1 and 91 (inclusive), which yields
.
See also
2006 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 8 |
Followed by Problem 10 | |
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.