Last two digit of the number (129)^129 will be

• 129^129
  (130 - 1)^129
  In bionomial expansion
  (a + b)^n
  (r +1)th term is given by
  t(r + 1) = nCr * a^(n-r) * b^r
  and has (n + 1)th term
  
  And I think that last two digits can be obtained by last two terms so lets see
  here
  a = 130, b = -1, n = 129
  t(129) = 129C128 * 130^(129 - 128) * (-1)^128
  t(129) = 129 * 130 * 1 = 16770
  
  t(130) = 129C129 * 130^(129 - 129) * (-1)^129
  t(130) = 1 * 1 * -1 = -1
  
  t(129) + t(130) = 16770 - 1 = 16769
  So
  last two digits are
  69
• 9 years agoHelpfull: Yes(1) No(1)
• (129)^129 = 9^129 = (9)^1*(9)^128 = 9*1 = 9 Answer
• 9 years agoHelpfull: Yes(0) No(3)
• (129)^129
  = 9^129
  = (9)^1*(9)^128
  =(-1+10)^128 * 9
  =128c0* (-1)^128 *10c0 + 128c1 * (-1)^127 * 10c1
  = 1-1280 = -1279
  Now
  = -1279*9
  =11511
  means last two digit will be 11
  option 2 is correct
• 9 years agoHelpfull: Yes(0) No(1)
• can u pls explain @jay kumar
• 9 years agoHelpfull: Yes(0) No(0)
• 129^129
  
  29 ^129
  (29^(2*64)+1) = 29^(2*64) . 29
  
  29^2 last 2 digit 41
  41^64 . 29
  41 ^ 64 last two digit 61
  61*29 = last two digit 69
• 9 years agoHelpfull: Yes(0) No(0)
• (129)^129
  = (100+29)^129/100 = (29)^129
  now, we can also write it as:
  (100-71)^129/100
  (-71)^129 = -39
  therefore, 100-39 = 61
  thus last two digit number is 61
  
  
  
  
  
  
  
  
  
• 9 years agoHelpfull: Yes(0) No(0)