2020 CIME I Problems/Problem 6
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. 
