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I first noticed that primes above 50 appear an even number of times, and at least the higher primes below 50 appeared an odd number of times. Therefore I figured it had to be 48, 50, or 52. I then figured of the three 50 was the most likely answer and confirmed it with code.
Then I thought there must be an easier way.
So if you think of this as 100*(99!)^2 * 98*(97!)^2... 2*(1!)^2 you can reduce it to
(some huge perfect square) * 100 * 98 * 96... * 2
Then you pull out a 2 from each factor and get 2^50*(50!). So yeah, exclude the 50! and you have a perfect square.
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