Number of zeroes at the end of a factorial
There is a problem with your solution for large numbers ( >= 5^13 to be exact) because mulFive=mulFive*5; will go negative as the int overflows. 

To protect against this, you could do 

 while(num/5!=0)
 {
     zeroCount=zeroCount+ (num/5);
     num = num/5;
 }


i.e. divide the number rather than multiply the divisor, then only the size of num is a limit (2^31-1)