Difference between revisions of "2018 AIME I Problems/Problem 6"
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+ | ==Problem== | ||
+ | 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>. | ||
+ | ==Solution 1== | ||
+ | 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>. | ||
+ | |||
+ | ==Solution 2== | ||
+ | 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>. | ||
+ | |||
+ | == 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 <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. | ||
+ | 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>) | ||
+ | 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. | ||
+ | Our total count is thus <math>1440</math>, yielding a final answer of <math>\boxed{440}</math>. | ||
+ | |||
+ | == Solution 4 == | ||
+ | |||
+ | 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>. | ||
+ | |||
+ | == Solution 5 == | ||
+ | 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>. | ||
+ | |||
+ | ==Video Solution== | ||
+ | |||
+ | https://www.youtube.com/watch?v=iE8paW_ICxw | ||
+ | |||
+ | == See also == | ||
+ | {{AIME box|year=2018|n=I|num-b=5|num-a=7}} | ||
+ | {{MAA Notice}} |
Revision as of 21:34, 23 July 2020
Contents
Problem
Let be the number of complex numbers with the properties that and is a real number. Find the remainder when is divided by .
Solution 1
Let . This simplifies the problem constraint to . This is true if . Let be the angle makes with the positive x-axis. Note that there is exactly one for each angle . This must be true for values of (it may help to picture the reference angle making one orbit from and to the positive x-axis; note every time ). For each of these solutions for , there are necessarily solutions for . Thus, there are solutions for , yielding an answer of .
Solution 2
The constraint mentioned in the problem is equivalent to the requirement that the imaginary part is equal to . Since , let , then we can write the imaginary part of . Using the sum-to-product formula, we get or . The former yields solutions, and the latter yields solutions, giving a total of solution, so our answer is .
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 . We have two cases to consider. Either , or and are reflections across the imaginary axis. If , then . Thus, or , giving us 600 solutions. (Equalities are taken modulo ) For the second case, . This means , giving us 840 solutions. Our total count is thus , yielding a final answer of .
Solution 4
Because we know that Hence Because is real, it is equal to its complex conjugate. Hence Substituting the expression we that we derived earlier, we get This leaves us with a polynomial whose leading term is Hence our answer is .
Solution 5
Since , let . For to be real, the imaginary parts of and must be equal, so . We need to find all solutions for in the interval . This can be done by graphing and and finding their intersections. Since the period of is and the period of is , the common period of both graphs is . Therefore, we only graph the functions in the domain . We can clearly see that there are twelve points of intersection. However, since we only graphed of the interval , we need to multiply our answer by . The answer is .
Video Solution
https://www.youtube.com/watch?v=iE8paW_ICxw
See also
2018 AIME I (Problems • Answer Key • Resources) | ||
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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