Find the number of zeros at the end of given problem

1!^1!*2!^2!*3!^3!...........up to 6!^6!

• no of zeros=pair of 5*2
  so 5!^5!=5^(5*4*3*2*1)=5^120 these are no of 5 in 5!*5!
  now in 6!^61=5^(6*5*4*3*2)=5^720 no of 5
  total no of 5=5^120 *5^720=5^840
  so total 840 zeros
• 9 years agoHelpfull: Yes(15) No(1)
• Here we should find out how many 5 are there
  So 5!^5! And 6!^6!
  =5^5!*5^6!
  =5^120+720
  =5^840
  840 are there
• 9 years agoHelpfull: Yes(8) No(1)
• no of zeros=pair of 5*2
  so 5!^5!=5^(5*4*3*2*1)=5^120 these are no of 5 in 5!*5!
  now in 6!^61=5^(6*5*4*3*2)=5^720 no of 5
  total no of 5=5^120 *5^720=5^840
  so total 840 zeros
  
• 9 years agoHelpfull: Yes(1) No(1)
• I dont think there would be any zeroes AT THE END, because the question is asking to find out the number of zeroes AT THE END, and not the number of zeroes in the end. the first 4 terms' unit digit sum dont count upto 0.
  
• 9 years agoHelpfull: Yes(1) No(0)