Difference between revisions of "Mock AIME 1 2010 Problems/Problem 6"
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== Solution == | == Solution == | ||
− | <math>\boxed{076}</math>. | + | Because <math>|z|<10000</math> and <math>z=w^4</math>, we know that <math>|w|<\sqrt[4]{10000}=10</math>. Thus, the number of Gaussian integers within the open [[disk]] that satisfies <math>|w|<10</math> is the number of possible <math>w</math>. This is equal to the number of [[lattice points]] in the open disk <math>x^2+y^2<100</math> on the <math>xy</math> plane. However, we have to be careful, because for every <math>z\neq0</math>, there are <math>4</math> different values of <math>w</math> which satisfy <math>z=w^4</math>. These <math>4</math> different values are all <math>i</math>-multiples of each other, so they are all Gaussian integers, and none of them lie in the same quadrant. Furthermore, because, from the problem, <math>w\neq z</math>, we know that <math>w</math> cannot be <math>0</math> or <math>1</math>. However, <math>z</math> can still equal <math>1</math> if <math>w=-1</math>, so <math>w\neq 1</math> will not affect our final answer. Thus, we will count the number of lattice points in the open disk <math>x^2+y^2<100</math> on the positive <math>x</math>-axis and the first quadrant. |
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+ | When <math>x=9</math>, <math>0\leq y\leq4</math>, so we have <math>5</math> points here. | ||
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+ | When <math>x=8</math>, <math>0\leq y\leq5</math>, so we have <math>6</math> points here (remember we are dealing with a [[strict inequality]], so <math>y=6</math> does not work). | ||
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+ | When <math>x=7</math>, <math>0\leq y\leq7</math>, so we have <math>8</math> points here. | ||
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+ | When <math>x=6</math>, <math>0\leq y\leq7</math>, so we have <math>8</math> points here (<math>y=8</math> does not work). | ||
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+ | When <math>x=5</math>, <math>0\leq y\leq8</math>, so we have <math>9</math> points here. | ||
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+ | When <math>x\in\{1,2,3,4\}</math>, <math>0\leq y\leq9</math>, so we have <math>10</math> points for each of these four values for a total of <math>40</math> points. | ||
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+ | We are not counting the points where <math>x=0</math>, because those are on the <math>y</math>-axis. Thus, our final answer is <math>5+6+8+8+9+40=\boxed{076}</math>. | ||
== See Also == | == See Also == | ||
{{Mock AIME box|year=2010|n=1|num-b=5|num-a=7}} | {{Mock AIME box|year=2010|n=1|num-b=5|num-a=7}} |
Latest revision as of 12:06, 2 August 2024
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 |