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>\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>.
+
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 $z$ satisfying $|z|=1$ and $z^{850}+z^{350}+1=0$.

Solution

We reduce the problem to $z^{17}+z^7+1$, remembering to multiply the final product by 50. We need the imaginary parts of the numbers $z^{17},z^7$ to cancel, which by working modulo 360 we can easily determine only happens when the number is of the form $cis(15x)$ (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 $cis(x)$ 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 $100$.

See also

2020 CIME I (ProblemsAnswer KeyResources)
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

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