How many trailing zeroes would be found in 4617!, upon expansion?

• for no of trailing zeros we have to find the no of 5 in 4617!,because 4617!=1*2*3*4*5*.........4616*4617.
  no of 5 in 4617!= (4617/5^1)+(4617/5^2)+(4617/5^3)+(4617/5^4)+(4617/5^5)+... up to 5^n 4617.
  now no of 5 =4617/5 + 4617/25 + 4617/125 + 4617/625 +4617/3125
  =923 + 184 + 36 + 7 + 1 ( only take the integer value )
  =1151
  so the no of zeros at the end of 4617! is 1151.
• 9 years agoHelpfull: Yes(27) No(3)
• 51 : 4617 ÷ 5 = 923.4, so I get 923 factors of 5
  52 : 4617 ÷ 25 = 184.68, so I get 184 additional factors of 5
  53 : 4617 ÷ 125 = 36.936, so I get 36 additional factors of 5
  54 : 4617 ÷ 625 = 7.3872, so I get 7 additional factors of 5
  55 : 4617 ÷ 3125 = 1.47744, so I get 1 more factor of 5
  56 : 4617 ÷ 15625 = 0.295488, which is less than 1, so I stop here.
  Then 4617! has 923 + 184 + 36 + 7 + 1 = 1151 trailing zeroes.
• 9 years agoHelpfull: Yes(7) No(2)
• @hemalatha we take 5 as it is starting number whose fact is 120 getting zero as the last digit
• 9 years agoHelpfull: Yes(1) No(0)
• y ua takin 5 ?? plz xplain
• 9 years agoHelpfull: Yes(0) No(0)
• explain it clearly how 923,184,36,7,1 came
  
• 9 years agoHelpfull: Yes(0) No(0)
• Trailing zeros obtained ....by the concept of highest power of a xyz.! when we divided a certain no.
  so in this prb t.z can be obtained by dividing by 10=5*2. so we take 5 has higest.
• 9 years agoHelpfull: Yes(0) No(0)
• yes 1151 is the ans
  we are dividing it by 5 bcz its a rule , wanna check out the explanation go through arun sharma
• 8 years agoHelpfull: Yes(0) No(0)