Difference between revisions of "2018 AIME I Problems/Problem 6"

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==Solution==
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==Problem==
Let <math>a=z^{120}</math>. This simplifies the problem constraint to <math>a^6-a \in \mathbb{R}</math>. This is true iff <math>Im(a^6)=Im(a)</math>. Let <math>\theta</math> be the angle <math>a</math> makes with the positive x-axis. Note that there is exactly one <math>a</math> for each angle <math>0\le\theta<2\pi</math>. This must be true for <math>12</math> values of <math>a</math> (it may help to picture the reference angle making one orbit from and to the positive x-axis; note every time <math>\sin\theta=\sin{6\theta}</math>). For each of these solutions for <math>a</math>, there are necessarily <math>120</math> solutions for <math>z</math>. Thus, there are <math>12*120=1440</math> solutions for <math>z</math>, yielding an answer of <math>\boxed{440}</math>.
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Let <math>N</math> be the number of complex numbers <math>z</math> with the properties that <math>|z|=1</math> and <math>z^{6!}-z^{5!}</math> is a real number. Find the remainder when <math>N</math> is divided by <math>1000</math>.
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==Solution 1==
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Let <math>a=z^{120}</math>. This simplifies the problem constraint to <math>a^6-a \in \mathbb{R}</math>. This is true if <math>\text{Im}(a^6)=\text{Im}(a)</math>. Let <math>\theta</math> be the angle <math>a</math> makes with the positive x-axis. Note that there is exactly one <math>a</math> for each angle <math>0\le\theta<2\pi</math>. This must be true for <math>12</math> values of <math>a</math> (it may help to picture the reference angle making one orbit from and to the positive x-axis; note every time <math>\sin\theta=\sin{6\theta}</math>). For each of these solutions for <math>a</math>, there are necessarily <math>120</math> solutions for <math>z</math>. Thus, there are <math>12*120=1440</math> solutions for <math>z</math>, yielding an answer of <math>\boxed{440}</math>.
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==Solution 2==
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The constraint mentioned in the problem is equivalent to the requirement that the imaginary part is equal to <math>0</math>. Since <math>|z|=1</math>, let <math>z=\cos \theta + i\sin \theta</math>, then we can write the imaginary part of <math> \Im(z^{6!}-z^{5!})=\Im(z^{720}-z^{120})=\sin\left(720\theta\right)-\sin\left(120\theta\right)=0</math>. Using the sum-to-product formula, we get <math>\sin\left(720\theta\right)-\sin\left(120\theta\right)=2\cos\left(\frac{720\theta+120\theta}{2}\right)\sin\left(\frac{720\theta-120\theta}{2}\right)=2\cos\left(\frac{840\theta}{2}\right)\sin\left(\frac{600\theta}{2}\right)\implies \cos\left(\frac{840\theta}{2}\right)=0</math> or <math>\sin\left(\frac{600\theta}{2}\right)=0</math>. The former yields <math>840</math> solutions, and the latter yields <math>600</math> solutions, giving a total of <math>840+600=1440</math> solution, so our answer is <math>\boxed{440}</math>.
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== Solution 3 ==
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As mentioned in solution one, for the difference of two complex numbers to be real, their imaginary parts must be equal. We use exponential form of complex numbers. Let <math>z = e^{i \theta}</math>. We have two cases to consider. Either <math>z^{6!} = z^{5!}</math>, or <math>z^{6!}</math> and <math>z^{5!}</math> are reflections across the imaginary axis.
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If <math>z^{6!} = z^{5!}</math>, then <math>e^{6! \theta i} = e^{5! \theta i}</math>. Thus, <math>720 \theta = 120 \theta</math> or <math>600\theta = 0</math>, giving us 600 solutions. (Equalities are taken modulo <math>2 \pi</math>)
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For the second case, <math>e^{6! \theta i} = e^{(\pi - 5!\theta)i}</math>. This means <math>840 \theta = \pi </math>, giving us 840 solutions.
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Our total count is thus <math>1440</math>, yielding a final answer of <math>\boxed{440}</math>.
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== Solution 4 ==
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Because <math>|z| = 1,</math> we know that <math>z\overline{z} = 1^2 = 1.</math> Hence <math>\overline{z} = \frac 1 {z}.</math> Because <math>z^{6!}-z^{5!}</math> is real, it is equal to its complex conjugate. Hence <math>z^{6!}-z^{5!} = \overline{z^{6!}}-\overline{z^{5!}}.</math> Substituting the expression we that we derived earlier, we get <math>z^{720}-z^{120} = \frac 1{z^{720}} - \frac 1{z^{120}}.</math> This leaves us with a polynomial whose leading term is <math>z^{1440}.</math> Hence our answer is <math>\boxed{440}</math>.
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== Solution 5 ==
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Since <math>|z|=1</math>, let <math>z=\cos \theta + i\sin \theta</math>. For <math>z^{6!}-z^{5!}</math> to be real, the imaginary parts of <math>z^{6!}</math> and <math>z^{5!}</math> must be equal, so <math>\sin 720\theta=\sin 120\theta</math>. We need to find all solutions for <math>\theta</math> in the interval <math>[0,2\pi)</math>. This can be done by graphing <math>y=\sin 720\theta</math> and <math>y=\sin 120\theta</math> and finding their intersections. Since the period of <math>y=\sin 720\theta</math> is <math>\frac{\pi}{360}</math> and the period of <math>y=\sin 120\theta</math> is <math>\frac{\pi}{60}</math>, the common period of both graphs is <math>\frac{\pi}{60}</math>. Therefore, we only graph the functions in the domain <math>[0, \frac{\pi}{60})</math>. We can clearly see that there are twelve points of intersection. However, since we only graphed <math>\frac{1}{120}</math> of the interval <math>[0,2\pi)</math>, we need to multiply our answer by <math>120</math>. The answer is <math>12 \cdot 120 = 1440 = \boxed{440} (mod 1000)</math>.
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==Video Solution==
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https://www.youtube.com/watch?v=iE8paW_ICxw
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== See also ==
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{{AIME box|year=2018|n=I|num-b=5|num-a=7}}
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{{MAA Notice}}

Revision as of 21:34, 23 July 2020

Problem

Let $N$ be the number of complex numbers $z$ with the properties that $|z|=1$ and $z^{6!}-z^{5!}$ is a real number. Find the remainder when $N$ is divided by $1000$.

Solution 1

Let $a=z^{120}$. This simplifies the problem constraint to $a^6-a \in \mathbb{R}$. This is true if $\text{Im}(a^6)=\text{Im}(a)$. Let $\theta$ be the angle $a$ makes with the positive x-axis. Note that there is exactly one $a$ for each angle $0\le\theta<2\pi$. This must be true for $12$ values of $a$ (it may help to picture the reference angle making one orbit from and to the positive x-axis; note every time $\sin\theta=\sin{6\theta}$). For each of these solutions for $a$, there are necessarily $120$ solutions for $z$. Thus, there are $12*120=1440$ solutions for $z$, yielding an answer of $\boxed{440}$.

Solution 2

The constraint mentioned in the problem is equivalent to the requirement that the imaginary part is equal to $0$. Since $|z|=1$, let $z=\cos \theta + i\sin \theta$, then we can write the imaginary part of $\Im(z^{6!}-z^{5!})=\Im(z^{720}-z^{120})=\sin\left(720\theta\right)-\sin\left(120\theta\right)=0$. Using the sum-to-product formula, we get $\sin\left(720\theta\right)-\sin\left(120\theta\right)=2\cos\left(\frac{720\theta+120\theta}{2}\right)\sin\left(\frac{720\theta-120\theta}{2}\right)=2\cos\left(\frac{840\theta}{2}\right)\sin\left(\frac{600\theta}{2}\right)\implies \cos\left(\frac{840\theta}{2}\right)=0$ or $\sin\left(\frac{600\theta}{2}\right)=0$. The former yields $840$ solutions, and the latter yields $600$ solutions, giving a total of $840+600=1440$ solution, so our answer is $\boxed{440}$.

Solution 3

As mentioned in solution one, for the difference of two complex numbers to be real, their imaginary parts must be equal. We use exponential form of complex numbers. Let $z = e^{i \theta}$. We have two cases to consider. Either $z^{6!} = z^{5!}$, or $z^{6!}$ and $z^{5!}$ are reflections across the imaginary axis. If $z^{6!} = z^{5!}$, then $e^{6! \theta i} = e^{5! \theta i}$. Thus, $720 \theta = 120 \theta$ or $600\theta = 0$, giving us 600 solutions. (Equalities are taken modulo $2 \pi$) For the second case, $e^{6! \theta i} = e^{(\pi - 5!\theta)i}$. This means $840 \theta = \pi$, giving us 840 solutions. Our total count is thus $1440$, yielding a final answer of $\boxed{440}$.

Solution 4

Because $|z| = 1,$ we know that $z\overline{z} = 1^2 = 1.$ Hence $\overline{z} = \frac 1 {z}.$ Because $z^{6!}-z^{5!}$ is real, it is equal to its complex conjugate. Hence $z^{6!}-z^{5!} = \overline{z^{6!}}-\overline{z^{5!}}.$ Substituting the expression we that we derived earlier, we get $z^{720}-z^{120} = \frac 1{z^{720}} - \frac 1{z^{120}}.$ This leaves us with a polynomial whose leading term is $z^{1440}.$ Hence our answer is $\boxed{440}$.

Solution 5

Since $|z|=1$, let $z=\cos \theta + i\sin \theta$. For $z^{6!}-z^{5!}$ to be real, the imaginary parts of $z^{6!}$ and $z^{5!}$ must be equal, so $\sin 720\theta=\sin 120\theta$. We need to find all solutions for $\theta$ in the interval $[0,2\pi)$. This can be done by graphing $y=\sin 720\theta$ and $y=\sin 120\theta$ and finding their intersections. Since the period of $y=\sin 720\theta$ is $\frac{\pi}{360}$ and the period of $y=\sin 120\theta$ is $\frac{\pi}{60}$, the common period of both graphs is $\frac{\pi}{60}$. Therefore, we only graph the functions in the domain $[0, \frac{\pi}{60})$. We can clearly see that there are twelve points of intersection. However, since we only graphed $\frac{1}{120}$ of the interval $[0,2\pi)$, we need to multiply our answer by $120$. The answer is $12 \cdot 120 = 1440 = \boxed{440} (mod 1000)$.

Video Solution

https://www.youtube.com/watch?v=iE8paW_ICxw

See also

2018 AIME I (ProblemsAnswer KeyResources)
Preceded by
Problem 5
Followed by
Problem 7
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
All AIME Problems and Solutions

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