Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

share|improve this question

1 Answer 1

up vote 36 down vote accepted

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 $199$.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.