Difference between revisions of "2020 CIME I Problems/Problem 6"
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| + | We reduce the problem to <math>z^17+z^7+1</math>, remembering to multiply the final product by 50. We need the imaginary parts of the numbers <math>z^17,z^7</math> to cancel, which by working modulo 360 we can easily determine only happens when the number is of the form <math>\cis(15x)</math>(this holds true because we are only looking for solutions with a magnitude of 1). We also need the real parts to sum to -1. We check all the multiples of 15 that result in <math>\cis(x)</math> being negative, and find that only two work(or alternatively, if you are good, you can guess that only 120 and 240 work). The answer is then 100. | ||
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{{CIME box|year=2020|n=I|num-b=5|num-a=7}} | {{CIME box|year=2020|n=I|num-b=5|num-a=7}} | ||
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Revision as of 17:30, 31 August 2020
Problem 6
Find the number of complex numbers 
satisfying 
and 
.
Solution
We reduce the problem to 
, remembering to multiply the final product by 50. We need the imaginary parts of the numbers 
to cancel, which by working modulo 360 we can easily determine only happens when the number is of the form $\cis(15x)$ (Error compiling LaTeX. Unknown error_msg)(this holds true because we are only looking for solutions with a magnitude of 1). We also need the real parts to sum to -1. We check all the multiples of 15 that result in $\cis(x)$ (Error compiling LaTeX. Unknown error_msg) being negative, and find that only two work(or alternatively, if you are good, you can guess that only 120 and 240 work). The answer is then 100.
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| 2020 CIME 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 CIME Problems and Solutions | ||
The problems on this page are copyrighted by the MAC's Christmas Mathematics Competitions. 
