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December 6th, 2015, 03:52 PM   #1
Joined: Jun 2015
From: Ohio

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Putnam Exam Question

Most of the questions from this year's Putnam exam completely stumped me, but here was one I was very curious about.


$\displaystyle \log_2 \prod_{a=1}^{2015} \prod_{b=1}^{2015} (1 + e^{2 \pi i ab/2015}) $

I had no idea how to do this. I thought I did at first, but then I remembered typical exponent laws don't work when you're working with complex numbers so I couldn't turn the $\displaystyle e^{2 \pi i}$ into 1 (since there were other things in the exponent). Anybody know how to do this?
numberguru1 is offline  
December 9th, 2015, 09:14 AM   #2
Joined: Oct 2013

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Hard for calculating a double prod up to 2015.
Even for a computer build in 2015!

We know:
2015 = 5*13*31

replace 2015 by 5 and calculate 9
replace 2015 by 13 and calculate 25
replace 2015 by 31 and calculate 61

finally, calculate the answer 9*25*61 = 13725

math is fun!
Martin Hopf is offline  
December 9th, 2015, 10:03 AM   #3
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I think it's just a case of counting the conjugate pairs.
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