Difference between revisions of "2006 AIME I Problems/Problem 9"
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Since <math>1003/11 = 91 + 2/11</math>, the answer is just the number of odd integers in <math>[1,91]</math>, which is <math>46</math>. | Since <math>1003/11 = 91 + 2/11</math>, the answer is just the number of odd integers in <math>[1,91]</math>, which is <math>46</math>. | ||
+ | ---------------- | ||
+ | == Solution 2 == | ||
+ | Using the above method, we can derive that <math>a^{2}r^{11} = 2^{1003}</math>. | ||
+ | Now, think about what happens when r is an even power of 2. Then <math>a^{2}</math> 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 <math>2^{1}</math>, <math>2^{3}</math>, <math>2^{5}</math> .... all work for r, until r hits <math>2^{93}</math>, when it gets greater than <math>2^{1003}</math>, so the greatest value for r is <math>2^{91}</math>. All that's left is to count the number of odd integers between 1 and 91 (inclusive), which yields <math>\boxed{046}</math>. | ||
== See also == | == See also == |
Revision as of 19:33, 29 December 2015
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
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
.
Another way is to write
Since , the answer is just the number of odd integers in
, which is
.
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.