Difference between revisions of "2020 CIME I Problems/Problem 6"
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==Solution== | ==Solution== | ||
− | 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> | + | 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 <math>1</math>). We also need the real parts to sum to <math>-1</math>. 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 <math>120</math> and <math>240</math> work). The answer is then <math>100</math>. |
==See also== | ==See also== | ||
{{CIME box|year=2020|n=I|num-b=5|num-a=7}} | {{CIME box|year=2020|n=I|num-b=5|num-a=7}} | ||
{{MAC Notice}} | {{MAC Notice}} |
Revision as of 17:58, 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 (this holds true because we are only looking for solutions with a magnitude of ). We also need the real parts to sum to . We check all the multiples of 15 that result in being negative, and find that only two work(or alternatively, if you are good, you can guess that only and work). The answer is then .
See also
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.