Difference between revisions of "2018 AIME I Problems/Problem 6"
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== Solution 5 == | == 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} | + | 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} \pmod{1000}</math>. |
==Video Solution== | ==Video Solution== |
Revision as of 18:01, 16 February 2021
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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