Difference between revisions of "2018 AIME I Problems/Problem 6"
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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. | 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. | ||
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. | 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>\boxed{ | + | Our total count is thus <math>1440</math>, yielding a final answer of <math>\boxed{440}</math>. |
== See also == | == See also == | ||
{{AIME box|year=2018|n=I|num-b=5|num-a=7}} | {{AIME box|year=2018|n=I|num-b=5|num-a=7}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 23:59, 19 April 2018
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 polar 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. For the second case, . This means , giving us 840 solutions. Our total count is thus , yielding a final answer of .
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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