[N] Question 4: number theory

So the number of trailing $0$ s depends on the number of pairs of $2\times5$ . Since there are more $2$ s than $5$ s, we are only concerned in finding the number of $5$ s.

So,

$1! \quad to\quad4!$ --- $no \quad5s$

$5! \quad to\quad9!$ --- $one\quad 5s$

$10! \quad to\quad14!$ --- $two \quad5s$

...

$20! \quad to\quad24!$ --- $four \quad5s$

$25! \quad to\quad29!$ --- ${\textbf{SIX}} \quad5s$

$30!$ --- ${\textbf{SEVEN}}\quad 5s$

Note that beyond $25!$ , it's pattern change cause $25={ 5 }^{ 2 }$

So the total number of $5$ s is $5\times(1+2+3+4+6)+7=\boxed{87}$