finding the smallest number $n$ such that $n!=n(n+1)(n+2)(n+3)$ What is the smallest number $n$ such that $n!=n(n+1)(n+2)(n+3)$?
How will I solve this type of problems?

Okay... Is there only one solution?  Well, yeah... because $n!$ increases by a factor of $(n+1)!/n! = n+1$ as $n$ increases by one, while $n(n+1)(n+2)(n+3)$ only increases by $(n+1)(n+2)(n+3)(n+4)/n(n+1)(n+2)(n+3) = (n+4)/n = 1 + 4/n$.  As $n!$ increases faster for $n > 2$ there is at most one solution.
(Unless there is a solution for $n = 1$ or $0$ which obviously there isn't.)
So.....  Let's start.
$n! = n(n+1)(n+2)(n+3) \iff (n-1)! = (n+1)(n+2)(n+3)$
So $(n-4)!(n-3)(n-2)(n-1) = (n+1)(n+2)(n+3)$
$(n+1)(n+2)(n+3) = n^3 + 6n^2 + 11n + 6$and $(n-1)(n-2)(n-3) = n^3 - 6n^2 + 11n - 6$
So $(n-4)!(n^3 - 6n^2 + 11n-6) = n^3 + 6n^2 + 11n + 6$
So $[(n-4)!-1](n^3 + 11n) =[(n-4)!+1](6n^2 + 6)$
$\frac{(n-4)! - 1}{(n-4)! + 1} = 6\frac{n^2+1}{n^3+11n}$
$\frac{(n-4)!+1}{(n-4)! + 1}- \frac{2}{(n-4)! + 1} = 6(\frac{n^2+11}{n^3+11n}- \frac{10}{n^3+11n})$.
$1 - idgybit = 6\frac 1 n - oodgybit$
So $n \approx 6$.
Iff $n=6$
$1 - \frac 2 3 = 1 - \frac{10}{36+11}$... Mmm.  not quite.
Iff $n = 7$
$1 - \frac 2{3! + 1} = 6/7 - \frac{60}{7^3 + 77}$
$1 - 2/7 = 6/7[1 - 10/(49 + 11)]$
$5/7 = 6/7[1 - 1/6] = 6/7*5/6 = 5/7$.  So that does it.
$n = 7$ is the only solution.
