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 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 |