2010 AIME I Problems/Problem 5
Contents
Problem
Positive integers , , , and satisfy , , and . Find the number of possible values of .
Solution 1
Using the difference of squares, , where equality must hold so and . Then we see is maximal and is minimal, so the answer is .
Note: We can also find that in another way. We know
Therefore, one of and must be Clearly, since then one would be positive and negative, or both would be zero. Therefore, so . Similarly, we can deduce that
Solution 2
Since must be greater than , it follows that the only possible value for is (otherwise the quantity would be greater than ). Therefore the only possible ordered pairs for are , , ... , , so has possible values.
See Also
2010 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 4 |
Followed by Problem 6 | |
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