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Difference between revisions of "Mock AIME 1 2010 Problems/Problem 6"

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== Solution ==
 
== Solution ==
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<math>\boxed{076}</math>.
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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 11:06, 2 August 2024

Problem

Find the number of Gaussian integers $z$ with magnitude less than 10000 such that there exists a different Gaussian integer $w$ such that $z = w^4$. (The magnitude of a complex $a+bi$, where $a$ and $b$ are reals, is defined to be $\sqrt{a^2+b^2}$. A Gaussian integer is defined to be a complex number whose real and imaginary parts are both integers.)

Solution

Because $|z|<10000$ and $z=w^4$, we know that $|w|<\sqrt[4]{10000}=10$. Thus, the number of Gaussian integers within the open disk that satisfies $|w|<10$ is the number of possible $w$. This is equal to the number of lattice points in the open disk $x^2+y^2<100$ on the $xy$ plane. However, we have to be careful, because for every $z\neq0$, there are $4$ different values of $w$ which satisfy $z=w^4$. These $4$ different values are all $i$-multiples of each other, so they are all Gaussian integers, and none of them lie in the same quadrant. Furthermore, because, from the problem, $w\neq z$, we know that $w$ cannot be $0$ or $1$. However, $z$ can still equal $1$ if $w=-1$, so $w\neq 1$ will not affect our final answer. Thus, we will count the number of lattice points in the open disk $x^2+y^2<100$ on the positive $x$-axis and the first quadrant.

When $x=9$, $0\leq y\leq4$, so we have $5$ points here.

When $x=8$, $0\leq y\leq5$, so we have $6$ points here (remember we are dealing with a strict inequality, so $y=6$ does not work).

When $x=7$, $0\leq y\leq7$, so we have $8$ points here.

When $x=6$, $0\leq y\leq7$, so we have $8$ points here ($y=8$ does not work).

When $x=5$, $0\leq y\leq8$, so we have $9$ points here.

When $x\in\{1,2,3,4\}$, $0\leq y\leq9$, so we have $10$ points for each of these four values for a total of $40$ points.

We are not counting the points where $x=0$, because those are on the $y$-axis. Thus, our final answer is $5+6+8+8+9+40=\boxed{076}$.

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
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