I originally started off by listing all the primes: $p<200$ then trying to calculate the prime factorisation of each (which I realise is a silly thing to do)

I believe there must be a simpler way to find the smallest and largest prime factors of $\dfrac{200!}{180!}$.

If I list the prime factorisation of $180$ and $200$ does that help me in any way?

I have calculated a similar thing before but with similar numbers and I'm not really sure how to deal with these larger numbers?

Thank you for any help


Note $\dfrac{200!}{180!}=200(199)\cdots(181)$. The smallest prime factor is easy to calculate - it's just $2$, since $2$ is the smallest prime and the product contains even factors. The largest prime factor is a little more interesting. You have to find the largest prime factor out of any of the elements in the product. However, noting that $199$ is prime, $200$ has no prime factors greater than $199$, and $199$ is greater than all other elements in the product - never mind their prime factors - gives us the result that the largest prime factor is $199$.


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