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 ![$|w|<\sqrt[4]{10000}=10$](http://latex.artofproblemsolving.com/e/1/0/e10c45d093d3d920ae18e34f29a209c0103d104b.png)
. 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 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
