If N! has 26 zeroes at the end, what is the maximum value of N?

• Ans:114
  Solution:
  For N!
  (N/5^1)+(N/5^2)+(N/5^3).....(only till N>denominator)=Number of zeros
  Assume N=110
  Therefore,(110/5)+(110/25)=22+4=26 Zeroes
  Number of Zeros is 26 upto 114!
  For 115!, number of zeroes is 27.
  So the maximum value for the given question is N=114.
  
• 8 years agoHelpfull: Yes(0) No(1)