Can you give me a big number with many divisors? Let $\tau(n)$ denote the number of divisors of $n$ and $p$ prime.
Today in high school I've tried to find a number that is not very big, but has a huge number of divisors (compared to it's size), i.e. $\tau(n)/n$ has to be as big as possible. 
Obviously, $0 \leq \tau(1) \leq 1$, since $\tau(1)/1=1$ and $\tau(p)/p=2/p$ (so this can become as small as we want). I have tried some values


*

*$\tau(2)/2 = 1$   

*$\tau(3)/3 = 2/3$ 

*$\tau(4)/4 = 0.75$ 

*$\tau(5)/5 = 0.4$


However, I am wondering whether there is a big(er) number, with a high ratio of divisors? Exists any number that e.g. has again a ratio of 2/3?
Such a number is not in the form $p^x$, since 
$$ \frac{\tau(p^x)}{p^x} = \frac{x+1}{p^x}, $$
which become smaller and smaller for bigger $p$ or $x$.
Also not $\prod^\omega_i p_i$, since
$$ \frac{\tau\left(\prod^\omega_i p_i\right)}{\prod^\omega_i p_i} = \frac{2^\omega}{\prod^\omega_i p_i} = \prod^\omega_i \frac{2}{p}, $$
which also (if I am correct) tends to 0 for bigger $\omega$.
Or does this already show that the ratio $\tau(n)/n$ can be only smaller not bigger for any higher number?
 A: 
However, I am wondering whether there is a big(er) number, with a high ratio of divisors? Exists any number that e.g. has again a ratio of 2/3?

Yes, $\tau(6)/6 = \frac{2}{3}$
Note that in the best case scenario, $n$ has divisors $1,2,...,\sqrt{n}$ together with their co-factors $\frac{n}{1},\frac{n}{2},...\frac{n}{\sqrt(n)}$, so that's at most $2*\sqrt{n}$ divisors.  This should allow you to figure out the upper bound to any number with a certain ratio.  (and this also tells you that the ratio, while it may go up or down from number to number,  will in general go down)
For example, for a ratio of $\frac{1}{2}$, you have an upper bound of 16, since for $n > 16$, $n$ has less than $n/2$ divisors.
If you try a few numbers, you thus see that 12 will be the last one with a ratio of $\frac{1}{2}$ or more
A: If $n = \prod p_i^{m_i}$ then $\tau (n) = \prod (m_i + 1)$
So we want $\frac {\tau(n)}{n} = \frac{\prod(m_i + 1)}{\prod p_i^{m_i}}$
The actual values of the prime factors are irrelevant to the number of divisors so to get a high ratio we want the primes to be as small as possible.  So $p_i$ should be the $i$-th prime.  $p_1=2;p_2 = 3;p_3 = 5; etc.$.
Which $p_i$ are raised to which power $m_i$ is irrelevent so to get a high ratio we want the high powers to go to the lowest primes so $m_1 \ge m_2 \ge m_3 ....$.
And obviously as we take higher powers $m_i$ the ratio of $\frac {m_i + 1}{p_i^{m_i}}$ is going to decrease radically.
That's not very precise but it gives a good frame work.  For $n = p^m$ the highest ratio will be $n=2^m$ and the ratio is $(m+1)/2^m$ so for $n = 2,4,8,16... $ we have the ratio is $2/2=1,3/4,4/8 = 1/2,5/16, 6/32...$ etc.
For $n = p^mq^k$ the highest ratio will be $n = 2^{m+a}3^m$ and the ratio is $m(m+a)/2^{m+a}3^m$.  So for $n = 6;12; 36;24;72;216 etc$; the ratio is $\frac 46;\frac 6{12};\frac 9{36}$ etc.
For no having three prime factors we have $n = 2*3*5; 2^2*3*5;2^2*3^2*5;2^3*3*5$ yields ratios of $\frac {8}{30};\frac{12}{60}; \frac{18}{180};\frac{16}{120} etc.$
Okay, that isn't precise but it's clear that those are the highest ratios.
$\tau(n) = 1$ only if $n = 2$; $\tau(n) = 3/4$ only if $n = 4$; $\tau(n)=2/3$ only if $n = 6$ and $\tau(n)=1/2$ only if $n =8,12$.
A: Well I don't know if such a big number exist or not but I can make a guess how we can get that.
The number of divisors of a number $n$ whose prime factorisation is $p_1^{a_1}.p_2^{a_2}......p_n^{a_n}$ are $(a_1+1).(a_2+1).(a_3+1).....(a_n+1)$. 
To get the highest ratio, we need to select smallest prime numbers. Further, we need to manipulate all the $a_i$ in such a way so that the number if factors become maximum.
