What is the last non zero digit of 96!

• If we try to compute 96!
  then
  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 i.e last digit is 6
• 11 years agoHelpfull: Yes(4) No(9)
• Lets say D(N) denotes the last non zero digit of factorial, then the algo says
  D(N)=4*D[N/5]*D(Unit digit of N)[If tens digit of N is odd]
  D(N)=6*D[N/5]*D(Unit digit of N)[If tens digit of N is even]; Where [N/5] is greatest Integer Function
  
  #Problem 18
  Find the last non zero digit of 26!*33!.
  
  Solution Scheme and Approach
  D(26)=6*D[26/5]*D(6)=6*D(5)*D(6)=6*2*2=4[D(5) means last non zero digit of 5!=120 which is 2, same for D(6)]
  D(33)=4*D[33/5]*D(3)=4*D(6)*D(3)=4*2*6=8
  
  Hence last non aero digit of 26!*33!=4*8=2
  Remember
  D(1)=1 D(2)=2 D(3)=6 D(4)=4 D(5)=2 D(6)=2 D(7)=4 D(8)=2 D(9)=8
• 11 years agoHelpfull: Yes(3) No(0)
• Z(n) denotes the last non zero digit in n!
  
  Z(n) = {4 if tens digit is odd and 6 if tens digit is even) *
  Z(n/5) * z(Units digit) therefore Z(96) = 4 * Z(96/5) * Z(6)
  = 4 * z(19) * last non zero digit of 6!
  = 4 * last non zero digit of 19! * last non zero digit of 6! again, 4 * z(19/5) * last non zero digit of 9! =192
  = 4 * last non zero digit of (192) * last non zero digit of (120)
  = 4 * 2 * 2
  = 16 (answer is the last digit ie 2)
  = 6
• 10 years agoHelpfull: Yes(3) No(0)
• can u explain me how it can be solved
• 11 years agoHelpfull: Yes(0) No(0)
• satyanaryana:any alternative way??
• 11 years agoHelpfull: Yes(0) No(0)
• 6! is 6*5*4*3*2*1=720
  so last non zero is 2
  is my correct????????????
• 11 years agoHelpfull: Yes(0) No(2)
• hey can any one explain simple logic behind it...satyanarayana ur logic not clear
• 11 years agoHelpfull: Yes(0) No(0)
• ANS : 8
  up to 9! we can write..1*2*3*4*..9 = ....8
  same way 11-19 we can write ...1*2*3*4*..9 (only unit dig) = 8
  this is similar 4 (21-29),(31-39)...(81-89).... so total is 8^9 = 8
  now consider only 10*20*30...90 = 8 (take only 10's dig.)
  so 8*8 = 64 .... no for 91-96... = 1*2*3*4*5*6 = ..2
  so the unit digit is 64*2 = 128 ... i.e. 8...
  
• 11 years agoHelpfull: Yes(0) No(2)
• let n=96
  A=96/5 =19
  B=96%5 =1
  (2^A)*A!*B!
  (2^19)*19!*1!
  last digits
  8*8*6*4
  ans=6
• 6 years agoHelpfull: Yes(0) No(0)
• n = 96. Expressing 96 in terms of 5a+b, we have
  
  96 = 5*19+1. i.e a=19 and b=1.
  
  L(96!) = Last Digit of [ 219 x L(19!) x L(1!) ]
  
  Calculating the last digit of 219. Using the technique explained in this post : Units digit of 219 => Units digit of 23 = 8
  
  Next lets calculate L(19!).
  Again expressing 19 in terms of 5a + b, we have
  
  19 = 5×3 + 4.
  
  Thus L(19!) = Last Digit of [ 23 x L(3!) x L(4!) ]
  
  Last digit of 23 = 8.
  
  Right most non zero digit of L(3!) = 3×2 = 6
  
  Last non zero digit of L(4!) = 4x3x2 = 4.
  
  Hence L(19!) = Last digit of [8 x 6 x 4] = 2
  
  Now, getting back to our original problem
  L(96!) = Last Digit of [ 219 x L(19!) x L(1!) ]
  
  Last digit of 219 = 8
  
  L(19!) = 2
  
  L(1!) = 1
  
  Thus L(96!) = Last Digit of [ 8 x 6 x 1 ] = 6
• 1 year agoHelpfull: Yes(0) No(0)