Mock AIME 1 2010 Problems/Problem 6
Problem
Find the number of Gaussian integers with magnitude less than 10000 such that there exists a different Gaussian integer such that . (The magnitude of a complex , where and are reals, is defined to be . A Gaussian integer is defined to be a complex number whose real and imaginary parts are both integers.)
Solution
Because and , we know that . Thus, the number of Gaussian integers within the open disk that satisfies is the number of possible . This is equal to the number of lattice points in the open disk on the plane. However, we have to be careful, because for every , there are different values of which satisfy . These different values are all -multiples of each other, so they are all Gaussian integers, and none of them lie in the same quadrant. Furthermore, because, from the problem, , we know that cannot be or . However, can still equal if , so will not affect our final answer. Thus, we will count the number of lattice points in the open disk on the positive -axis and the first quadrant.
When , , so we have points here.
When , , so we have points here (remember we are dealing with a strict inequality, so does not work).
When , , so we have points here.
When , , so we have points here ( does not work).
When , , so we have points here.
When , , so we have points for each of these four values for a total of points.
We are not counting the points where , because those are on the -axis. Thus, our final answer is .
See Also
Mock AIME 1 2010 (Problems, Source) | ||
Preceded by Problem 5 |
Followed by Problem 7 | |
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