1000! contains how many trailing zeros? Could anybody explain how to find that in any simple way??

• 1000/5 +1000/25 +1000/125 +1000/625= 200+40+8+1= 249
  
  249 trailing zeros in 1000 !.
• 12 years agoHelpfull: Yes(5) No(2)
• Found Solution : 249 trailing zeros in 1000!
  
  Explanation : (May be clear to someone)
  
  In factorial we get Zero only when we get multiplied by 5 or multiples of 5.
  Therefore,
  From 1 to 1000, multiples of 5 = 1000/5 = 200 times (5,10,15,20,.......,995,1000)
  From 1 to 1000, multiples of 5^2 = 1000/25 = 40 times (25,50,75,......,975,1000)
  From 1 to 1000, multiples of 5^3 = 1000/125 = 8 times (125,250,....,875,1000)
  From 1 to 1000, multiples of 5^4 = 1000/625 = 1 time only (625 only)
  
  After that we not able to find multiples of 5. Because 5^5 = 3125(exceeds 1000)
  
  Now, we add all = 200+40+8+1 = 249 trailing zeros in 1000!
  
  :-)
• 12 years agoHelpfull: Yes(3) No(6)
• 249 trailing zeros are there.
• 12 years agoHelpfull: Yes(3) No(1)
• 1000/5 +1000/25 +1000/125 = 200+40+8= 248
  
  248 trailing zeros in 1000 !.
• 12 years agoHelpfull: Yes(1) No(1)