123^123! what is the last 2 digits?

• Last two digits are 01
  
  It can be find out using binomial theorem :-
  
  123^(123)! = ((123)^2)^(123!)/2
  
  123^2 = 15129
  last two digits are 29
  
  So we can write it as:-
  
  ((123)^2)^(123!)/2 = (-1+30)^(123!)/2
  
  Now (123!)/2, which is a very large number and will have many trailing zeroes
  
  we can expand the binomial expression as :-
  
  nC0 * a^n * b^0 + nC1 * a^(n-1)*b^1 + nC1 * a^(n-1)*b^1 .... ....(no need to go further for this question)
  
  here n = (123!)/2 , a= -1 b = 30
  
  Putting values:-
  
  (123!)/2 C0 * (-1)^(123!)/2 30^0 + (123!)/2 C1 * (-1)^((123!)/2 -1) * 30^1
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  last digit: 1 + last digit: 000000 + last digit: 000000
  
  So final addition will be like .....000001
  
  Last two digit 01 Ans
  
  jisko samajh nhi aya wo call krlo muje 09981241341, smjha dunga :P
  
  
• 11 years agoHelpfull: Yes(9) No(3)
• ans 67
  123^123 convert it such that its last digit is 1
  123^120*123^3=(123^4)^30*123^3=(41)^30(last 2 digits)*123^3
  41^30 last 2 digit= 4*0=0 & 1 i.e. 01 *123^3=67 ans
  
• 11 years agoHelpfull: Yes(6) No(5)
• last two digits: 01.
  
  explanation : 1. check the cyclicity of 123 that is 4
  123^123! = (123^4)^xyz000000....
  therefore 01
• 11 years agoHelpfull: Yes(5) No(8)
• (123^4)^xy000.. nw apply cyclicity rule answer will be 01
• 11 years agoHelpfull: Yes(1) No(2)
• ans is 67...
  
• 11 years agoHelpfull: Yes(0) No(5)
• since last 2 digits are needed so:
  123^123! mod 100
  Carmichael Number for 100 is 20
  Since 20 divides 123!, so answer is to be unity
  Hence 01
• 11 years agoHelpfull: Yes(0) No(0)
• 123^..........0000...28 zeroe's
  
  ==01 as last two digits !!!
• 11 years agoHelpfull: Yes(0) No(0)