A number theory problem by Dhrubajyoti Ghosh

There are $86$ multiples of $23$ less than $2000$ .

However, $529=23\cdot23, 1058=23\cdot23\cdot2, 1587=23\cdot23\cdot2\cdot2$ .

So there are total $89$ $23$ 's in $2000!$ .

So greatest positive integer x such that $23^{x+6}$ divides $2000!$ can be found as follows:

$x+6=89$

$\implies x=\boxed{83}$