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I have a problem with this question

John writes the number 1!, 2!, 3!, ... , 199!, 200! on a whiteboard. John then erases one of the numbers. John then multiplied the remaining 199 numbers. He found out that the number was a perfect square. What was the number that was erased?

Can someone explain?

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    $\begingroup$ You can prime-factorize 1!2!...200! to see which prime has an odd index. $\endgroup$
    – Kenny Lau
    Jun 25, 2016 at 3:25

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To be a perfect square, every prime factor has to appear an even number of times. $199$ is prime and is a factor of $199!$ and $200!$, so it appears an even number of times in the original product. If you erased $199!$ or $200!$ the product could not be square as it would have an odd number of $199$s. $197$ is prime. Which factorials does it appear in? It also restricts which factorials can be erased. What prime is the largest one that appears an odd number of times in the product? Are there others? You might be interested in De Polignac's formula for the exact power of a prime dividing a factorial. It is also known as Legendre's formula if you are searching.

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  • $\begingroup$ So since all prime numbers after two are odd numbers, which means they appear in an even number throughout 1! to 200!, then that means 2 is the only number which appears an odd number of times throughout 1! to 200!, right? So does that means 2 is the answer? $\endgroup$
    – Aden
    Jun 25, 2016 at 3:58
  • $\begingroup$ It's not that simple. All the primes above $100$ appear an even number of times for that reason. $97$ appears twice in $194!$ and above. You need to think about that. Then primes below $13$ are different because $13^2 \lt 200$ You need to think about all these cases. $\endgroup$
    – Ross Millikan
    Jun 25, 2016 at 4:04
  • $\begingroup$ Could you show me your working for this question? I think I could understand better through workings. $\endgroup$
    – Aden
    Jun 25, 2016 at 6:29

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