2006 AIME I Problems/Problem 9
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 .
Solution 3
Using the method from Solution 1, we get .
Since and both have to be powers of , we can rewrite this as .
. So, when we subtract from , the result is divisible by . Evaluating that, we get as a valid solution. Since , when we add to the value of , we can subtract from the value of to keep the equation valid. Using this, we get . In order to count the number of ordered pairs, we can simply count the number of values. Every odd number from to is included, so we have solutions.
-Phunsukh Wangdu
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 | ||
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