$\tau( n!(n+1) ) = 2\times\tau(n!)$. Then what is $n!\;\text{mod} \; (n+1)?$ 
If the number of factor of $(n+1)!$ is double than the number of factor of $n!$, then find the reminder if $n!$ is divided by $(n+1)$? 

I'm not sure if the question mean factor = divisor. However in both cases I cant find a way to start.
Source: BdMO 2015 Noril regional Higher Secondary Question no. 10. 
 A: According to the below theorem, $n$ as in the problem statement is either one less than a prime, or $n=3$. In the former case, Wilson's theorem gives us $n!\equiv n\pmod {n+1}$; in the latter case $3!\equiv 2\pmod 4$.
Theorem.
For a natural number $n$, we have $\tau(n!)=2\tau((n-1)!)$ iff $n$ is prime or $n=4$.
Proof.
One direction is straightforward: We have $\tau(4!)=8$, $\tau(6)=4$, and if $n$ is prime, the divisors of $n!$ are precisely the numbers $d$ and $nd$ where $d$ is a divisor of $(n-1)!$.
For the other direction, let  $n=p_1^{a_1}p_2^{a_2}\cdots p_m^{a_m}$ be the prime factorization of $n$ (with $m\ge1$ and all $a_i\ge 1$).
Write $n!=p_1^{b_1}p_2^{b_2}\cdots p_m^{b_m}r$ with $r$ only divisible by other primes (i.e., co-prime to $n$).
So $(n-1)!=p_1^{b_1-a_1}p_2^{b_2-a_2}\cdots p_m^{b_m-a_m}r$.
Then by multiplicativity of the divisor counting function,
$$\tau(n!)=(b_1+1)(b_2+1)\cdots (b_m+1)\tau(r)$$
and
$$ \tau((n-1)!)=(b_1-a_1+1)(b_2-a_2+1)\cdots (b_m-a_m+1)\tau(r).$$
Hence for $\tau(n!)=2\tau((n-1)!)$ we need
$$ 2\cdot (b_1-a_1+1)(b_2-a_2+1)\cdots (b_m-a_m+1)= (b_1+1)(b_2+1)\cdots (b_m+1)$$
or equivalently
$$\tag1 \prod_{i=1}^m\left(1-\frac{a_i}{b_i+1}\right)=\frac12.$$
It is well known that the exponent $b_i$ of $p_i$ in $n!$ is given by
$$\tag2 b_i=\left\lfloor \frac {n}{p_i}\right\rfloor+\left\lfloor \frac {n}{p_i^2}\right\rfloor+\left\lfloor \frac {n}{p_i^3}\right\rfloor+\ldots.$$
Let $Q_i=\frac{n}{p_i^{a_i}}$. 
Then for $1\le k\le a_i$, the $k$th summand in $(2)$ equals $p_i^{a_i-k}Q_i$, which gives us the simple estimates
$$\tag3b_i\ge a_iQ_i$$
and
$$\tag4b_i\ge (a_i-1)p_iQ_i+Q_i.$$ 
As $Q_i$ has $m-1$ distinct prime divisors, we conclude $Q_i\ge m!$.
Using this, $(1)$, $(3)$, and Bernoulli's inequality, we obtain
$$\tag5\begin{align}\frac12&=\prod_{i=1}^m\left(1-\frac{a_i}{b_i+1}\right)\\
& \ge \prod_{i=1}^m\left(1-\frac{a_i}{Q_ia_i+1}\right)\\
&>\prod_{i=1}^m\left(1-\frac{1}{Q_i}\right) \\
& \ge \left(1-\frac1{m!}\right)^m\\
&\ge 1-\frac 1{(m-1)!}.\end{align}$$
This leads to $(m-1)!<2$, i.e., $m\le2$.
Assume $m=2$.
Then from $(5)$, we extract
$$\frac12 > \left(1-\frac1{Q_1}\right)\left(1-\frac1{Q_2}\right).$$
If none of the $Q_i$ equals $2$, this leads to the absurdity $\frac12>\frac23\cdot\frac34$ because $Q_1\ne Q_2$.
Therefore, wlog. $Q_2=2$ and consequently $p_1=2$, $a_1=1$, $n=2p_2^{a_2}$ with $p_2\ge 3$, 
and $b_1\ge p_2^{a_2}+\lfloor p_2^{a_2}/2\rfloor \ge 4$.
Also, from $(4)$, we get $b_2\ge 6a_2-4$.
Plugging this into $(1)$, we arrive at
$$ \frac12 = \left(1-\frac1{b_1+1}\right)\left(1-\frac{a_2}{b_2+1}\right)\ge \frac 45\cdot\left(1-\frac{a_2}{6a_2-3}\right)\ge\frac45\cdot\frac 23,$$
contradiction.
Therefore,  $m=1$,  $Q_1=1$,  and as $(1)$ demands $2a_1=b_1+1$, we have from $(4)$ that
$$2(a_1-1) = b_1-1 \ge (a_1-1)p_1$$
and conclude that  $a_1=1$ (i.e., $n$ is prime) or $p_1=2$.
In the latter case, $b_1=2^{a_1}-1$, and $2a_1=b_1+1$ leads to $a_1\le 2$, i.e., we get only $n=4$ as additional possibility. 
$\square$
