2007 AMC 10A Problems/Problem 23
Problem
How many ordered pairs of positive integers, with , have the property that their squares differ by ?
Solution 1
For every two factors , we have . Since , , from which it follows that the number of ordered pairs is given by the number of ordered pairs . There are factors of , which give us six pairs . However, since are positive integers, we also need that are positive integers, so and must have the same parity. Thus we exclude the factors , and we are left with four pairs .
Solution 2
Similar to the solution above, reduce to . To find the number of distinct factors, add to both exponents and multiply, which gives us factors. Divide by since must be greater than or equal to . We don't need to worry about and being equal because is not a square number. Finally, subtract the two cases above for the same reason to get .
Solution 3
Find all of the factor pairs of 96: You can eliminate because you cannot have two numbers add to be an even number and have an odd difference at the same time without them being a decimal. You only have 4 left, so the answer is .
~HelloWorld21
See also
2007 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 22 |
Followed by Problem 24 | |
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All AMC 10 Problems and Solutions |
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