Find $n$ if $n!$ ends in exactly $100$ zeroes and $n$ is divisible by $8$. This question was in school maths challenge.
I dont know how to approach this one.. any help would be appreciated.

Ends in 100 zeros means that $5^{100}$ is a divisor.  If $n \ge 5^im$ then $n > 2^im$ so if $5^{100}$ is a divisor so is $2^{100}$ so is $10^{100}$. so $5^{100}$ divisor is sufficient to end in 100 zeros. 
To end in exactly 100 zeros means that in {1,2, ...., n} there are precisely 100 occurrences of 5 and added powers of 5.  Let  $5M \le n  <= 5(M+1)$.  There are M mulitiples of 5.  1/5 of those will be multiples of 25 and 1/25 will be multiples of 125 etc.
In total $5^m$ will yield $5^{m-1}$ multiples of 5, $5^{m-2}$ multiples of 25, etc.  So $5^m!$ will have $5^{m-1} + 5^{m-2} + 5^{m - 3}+ ....$ zeros. So 125! will have 25 multiples of 5, 5 multiples of 25,, and 1 multiple of 125 so 125! ends with exactly 31 zeros.  
So 375! will end with exactly 93 zeros.  (As there are 75 multiples of 5, 15 multiples of 25, and 3 multiples of 125).  We need 7 more zeros.  that is 6 more multiples of 5 and one more multiple of 25.  So 405! has exactly 100 zeros.  (81 multiples of 5, 16 multiples of 25 and 3 multiples of 125).
So the $405 \le n < 410$.  As 8|n.  n = 408. 
