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What is the largest prime factor of $\tau (20!)$ (where $\tau (n)$ is the number of divisors of $n$).

This question arises in a chapter of my number theory notes where the author shows that $v_{p}(n) = \lfloor (n/p) \rfloor + \cdots \lfloor(n/p^s) \rfloor$ where $p^{s+1} > n$, and $p$ is a prime number.

Further if a number $k = p_1^{e_1}\cdots p_k^{e_k}$ then I know that $\tau(k)=(e_1+1)\cdots (e_k+1)$, and that $\tau$ is mutiplicative.

Now I want to connect these ideas somehow to solve this problem. Hints appreciated.

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  • $\begingroup$ Hint: factoring $20!$ is the way to go (it is, perhaps, easier than you think!) $\endgroup$ – Lynn Apr 19 at 23:11
  • $\begingroup$ What is the number of divisors of divisors of $24$, i.e. $\tau(4!)$? Is it $8$ since the divisors are $1,2,3,4,6,8,12,24$, or is it more with some counted more than once? $\endgroup$ – Henry Apr 19 at 23:11
  • $\begingroup$ @Henry I think it's 8. I'm not sure what you mean by this but I want to first figure out what $\tau(20!)$ is, so your hint seems to be what I am trying to figure out. $\endgroup$ – IntegrateThis Apr 19 at 23:21
  • $\begingroup$ You said "number of divisors of divisors" rather than "number of divisors". $\endgroup$ – Henry Apr 19 at 23:30
  • $\begingroup$ @Henry my mistake, edited. $\endgroup$ – IntegrateThis Apr 19 at 23:31
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For $n=\prod p_i^{\alpha_i},\ \tau(n)=\prod(\alpha_i+1)$

As it happens, the largest exponent ($\alpha_i$) in the prime factorization of any factorial will be associated with $2$. For $20!$, the number of factors of $2$ is $18$, so $\alpha_i+1=19$ which happens to be prime. The other exponents associated with larger primes will all be smaller than $18$, so the largest prime factor of $\tau(20!)=\prod(\alpha_i+1)$ will be $19$ without having to do any further computations.

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So just calculate $v_p(20!)$ for the candidate primes $p=2,3,\ldots,19$ and find the largest prime occuring as a factor of some $v_p(20!)+1$.

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