Find the last non zero digit of 96!.
a) 2 b) 4 c) 6 d) 8

• it will be 6.
  ==4*z(96/5)*z(6)
  =4*z(19)*8
  =8*(4*z(19/5)*z(9))
  =8*(4*z(3)*8)
  =8*(4*6*8)
  =1536
  here we consider the unit digit of the result
• 11 years agoHelpfull: Yes(8) No(6)
• 2 is the right answer
  see this
  http://www.pagalguy.com/news/ive-got-power-working-with-factorials-cat-2011-quant-a-17445
• 11 years agoHelpfull: Yes(7) No(4)
• ans is 4, coz after 4! every factorial value ends in 0. eg 5! is 120, 6! is 720, 7! is 5040.. likewise upto n factorial value ends only in 0, the last factorial which has non zero digit is 4! tat is 24. so 4 is the answer..
• 11 years agoHelpfull: Yes(3) No(4)
• 4 is the answer
• 11 years agoHelpfull: Yes(2) No(1)
• 8 i think ans
• 11 years agoHelpfull: Yes(1) No(3)
• @udit: plz solve the question using the method specified in the link. i m gettng the answer as 6
• 11 years agoHelpfull: Yes(1) No(2)
• it will be 6.
  ==4*z(96/5)*z(6)
  =4*z(19)*8
  =8*(4*z(19/5)*z(9))
  =8*(4*z(3)*8)
  =8*(4*6*8)
  =1536
• 11 years agoHelpfull: Yes(0) No(5)
• @udit: but in this link how did we get 8 in the first example of 37!
• 11 years agoHelpfull: Yes(0) No(1)
• n=96 n=5a+b 5*19+1
  a=19,b=1
  last digit of [2+a x r(a!) * r(b!)]
  
• 11 years agoHelpfull: Yes(0) No(0)
• the answer should be 8
  96!= 5(19)+1
  where a=19 and b=1
  so by using : last digit of (2a * R(a!) * R(b!))
  therefore, (219*R(19!)*R(1!)) = 219*2*1= 8
  where R(19!) has the last non zero digit 2, it can b calculated by the same formula.
• 11 years agoHelpfull: Yes(0) No(0)
• answer is 6
• 8 years agoHelpfull: Yes(0) No(0)