2006 AIME A Problems/Problem 9
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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 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 .
See also
2006 AIME II (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 |