2000 AIME I Problems/Problem 6
Because , we only consider .
For simplicity, we can count how many valid pairs of that satisfy our equation.
The maximum that can be is because must be an integer (this is because , an integer). Then , and we continue this downward until , in which case . The number of pairs of , and so is then .
Let = and =
This makes counting a lot easier since now we just have to find all pairs that differ by 2.
Because , then we can use all positive integers less than 1000 for and .
Without loss of generality, let's say .
We can count even and odd pairs separately to make things easier*:
This makes 499 odd pairs and 498 even pairs, for a total of pairs.
Note: We are counting the pairs for the values of and , which, when squared, translate to the pairs of we are trying to find.
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