what is the last non-zero digit of 96!?

• Rightmost non-zero digit of n!
  R(n!)=last digit of [2^a. R(a!).R(b!)]
  
  where n=5a+b
  so 96=5*19+1
  i.e. a=19 and b=1
  
  last digit of [2^19. R(19!).R(1!)]-------(1)
  now again break 19 in the form of 5a+b
  19=5*3+4
  last digit of [2^3. R(3!).R(4!)]
  i.e. 2
  and put the value in (1)
  ans : 6
• 11 years agoHelpfull: Yes(22) No(4)
• 1*2*3*4*5*6*7*8*9*10
  last non zero digit=8
  11*12*13*.........*20
  last non zero digit=8
  21*.....*30=8(last non zero digit=8)
  31*.....*40=8(last non zero digit=8)
  41*.....*50=8(last non zero digit=8)
  51*.....*60=8(last non zero digit=8)
  61*.....*70=8(last non zero digit=8)
  71*.....*80=8(last non zero digit=8)
  81*.....*90=8(last non zero digit=8)
  91*92*..*96=8(last non zero digit=2)
  so=8^9*2
  last digit=6
• 11 years agoHelpfull: Yes(17) No(2)
• there is a formula for getting last non-zero digit of n!
  R(n!)=last digit of[2^a*R(a!)*R(b!)]
  where n=5*a+b.
  here n=96=5*19+1.
  so, a=19&b=1
  R(96!)=last digit of[2^19*R(19!)*R(1!)]
  =last digit of[(2^4)^4*2^3*2*1]
  =last digit of[6*8*2*1]
  =last digit of[36]
  =6.
  
  
  
  last nonzero digit of 19!
  19=5*3+4.
  a=3,b=4
  R(19!)=last digit of[2^3*R(3!)*R(4!)]
  =8*6*4
  =2.
  
• 11 years agoHelpfull: Yes(7) No(1)
• 1*2*3*4*5*6*7*8*9*10
  last non zero digit=8
  11*12*13*.........*20
  last non zero digit=8
  21*.....*30=8(last non zero digit=8)
  31*.....*40=8(last non zero digit=8)
  41*.....*50=8(last non zero digit=8)
  51*.....*60=8(last non zero digit=8)
  61*.....*70=8(last non zero digit=8)
  71*.....*80=8(last non zero digit=8)
  81*.....*90=8(last non zero digit=8)
  91*92*..*96=2(last non zero digit=2)
  so 8^9*2=6
  last digit=6
• 11 years agoHelpfull: Yes(6) No(0)
• @SUMAN KUMAR
  by having 91*92*93.....*96 = we have last non zero digit as 8, then how you considered it as 2.
  
• 11 years agoHelpfull: Yes(2) No(0)
• last nonzero digit of 96!=last digit of[Right most digit of 2^19*right most digit of 19!* right most digit of 1!]
  now right most nonzero digit of 19!=[right most digit of 2^3*right most digit of 3!*right most digit of 4!]=right most of[8*6*4]=2
  
  So,last nonzero digit of 96! is last digit of 8*2*1=16..i.e is 6 is the answer
• 11 years agoHelpfull: Yes(0) No(0)
• how u getting the values of A & B ?
• 11 years agoHelpfull: Yes(0) No(0)