upto how many integers we have to multiply 1*2*3*4*5*......
to make 31 zeros at the end

• we can do following
  if the number is close to multiple of 5
  for example
  take 101!
  the closest multiple is 100
  divide 100 by 5 we get 20 multiple, now we divide 100 by 5^2 we get 4 multiple
  thus we have 24 trailing zeros in 101 !
  similarly, for 125!
  125 divide by 5= 25 multiple
  125 divide by 5^2=5 multiple
  125 divide by 5^3=1 multiple
  thus total we have 31 trailing zeros
  in short, for trailing zeros
  n!= n/5+n/(5^2)+ n/(5^3)+.....
• 10 years agoHelpfull: Yes(1) No(1)
• the integers should multiply and give product as 125
  so
  125/5=25
  25/5=5
  5/5=1
  25+5+1=31
  31 zeroes will be there
• 10 years agoHelpfull: Yes(0) No(0)