Factorial

Giventhat n!，in decimal notation，has exactly 57 ending zeros，find the sum of
all possible values of n.
Sol
[240/5]+[240/25]+[240/12]=48+10+2=60>57
[239/5]+[239/25]+[239/125]=47+9+1=57
[238/5]+[238/25]+[238/125]=47+9+1=57
[237/5]+[237/25]+[237/125]=47+9+1=57
[236/5]+[236/25]+[236/125]=47+9+1=57
[235/5]+[235/25]+[235/125]=47+9+1=57
[234/5]+[234/25]+[234/125]=46+9+1=56<57
So
the sum of all possible values of n.
=239+239+237+236+235
=(239+235)*5/2
=1185