# What is the largest prime factor of $\tau (20!)$

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.

• Hint: factoring $20!$ is the way to go (it is, perhaps, easier than you think!) – Lynn Apr 19 at 23:11
• 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? – Henry Apr 19 at 23:11
• @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. – IntegrateThis Apr 19 at 23:21
• You said "number of divisors of divisors" rather than "number of divisors". – Henry Apr 19 at 23:30
• @Henry my mistake, edited. – IntegrateThis Apr 19 at 23:31

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.
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 $$v_p(20!)+1$$.