Find the number of zeroes in 58! (58 factorial)

• no. of zeroes in n! = [n/5]+[n/5^2]+[n/5^3]+...
  
  number of zeroes in 58! = [58/5]+[58/5^2]+[58/5^3]= 11+2+0 = 13
• 10 years agoHelpfull: Yes(22) No(3)
• no of zeroes is
  58/5=11
  11/5=2
  11+2=13
  13 is answer
  
• 10 years agoHelpfull: Yes(18) No(0)
• 58/5+58/25=11+2=13
• 10 years agoHelpfull: Yes(2) No(0)
• 58/5+58/5^2
  Taking quotient
  11+2=13
  So no.of zeros is 13
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• [58/5]+[58/5^2]=11+2=13
  13 zeroes exists in 58!
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• 58/5=11(quotient)
  11/5=2
  11+2=13(no.of zeros)
  
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• 13 zeros
  find out no of fives and no of two.
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• no of zeroes is
  58/5=11
  11/5=2
  11+2=13
  13
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• 58/5 quotient is 11
  11/5 quotient is 2
  number of zeroes = 11+2 =13
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• 13 since number of fives are 13 and 5 * even number will result in a zero.
  Thus it has 13 zeroes.
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• 58/5+58/25=13. (add only the quotient)
• 7 years agoHelpfull: Yes(0) No(0)