Find the number of zero’s in 173! (173 Factorial)

• no of zeroes in 173! is
  173/5=34(quotient)
  34/5=6(quotient)
  6/5=1(quotient)
  34+6+1=41
  ans 41 zeroes
• 10 years agoHelpfull: Yes(53) No(0)
• no. of zeroes in n! = [n/5]+[n/5^2]+[n/5^3]+...
  
  number of zero’s in 173! = [173/5]+[173/5^2]+[173/5^3]+[125/5^4]= 34+6+1+0 = 41
• 10 years agoHelpfull: Yes(30) No(6)
• Ans:41
  NO of zero’s in 173! = [173/5]+[173/5^2]+[173/5^3]+[125/5^4]= 34+6+1+0
  
  
• 10 years agoHelpfull: Yes(2) No(1)
• 173/5+173/25+173/125=34+6+1=41
• 10 years agoHelpfull: Yes(2) No(0)
• If you have count the number of zero's in n!
  n/5+n/(5^2)+n/(5^3)...
  173/5+173/25+173/125=34+6+1=41
• 9 years agoHelpfull: Yes(1) No(0)
• 173!=173%5+173%25+175%125=34+6+1=41
  
• 9 years agoHelpfull: Yes(1) No(0)
• 173/5+173/25+173/125
  Taking the quotient
  34+6+1=41
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• 173/5=34
  173/25=6
  173/125=1
  34+6+1=41
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• 173/5=34
  34/5=6
  6/5=1
  so 34+6+1=41 zeroes
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• 173/5+173/25+173/125
  Taking the quotient
  34+6+1+0=41
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• 52 no of zeros.
• 9 years agoHelpfull: Yes(0) No(6)
• 173/5 quotient=34
  34/5 quotient=6
  6/5 quotient=1
  hence the number of zeroes=34+6+1=41
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• 41
  10=2*5
  (173/5)+(173/25)+(173/125)=34+6+1=41
  
• 9 years agoHelpfull: Yes(0) No(0)