Find the last 2 digits of 994^499.

• 994 (mod 100) = 94 (mod 100).
  Please note Euler phi of 100 = 100*(1-1/2)*(1-1/5) = 40
  So, we take power module 40, 499(mod 40) = 19 (mod 40).
  Now, the original problem becomes 94^19 (mod 100)
  = (4+90)^19 (mod 100)
  = 4^19 + 19*90*4^18 (mod 100) by using binomial theorem. Rest terms are ignored because those are divisible by 100.
  = 4^19 + 10*4^18 (mod 100)
  
  4^19 (mod 100) = 44 (mod 100)
  4^18 (mod 10) = 6 (mod 10)
  So, 4^19 + 10*4^18 (mod 100) = (44 + 10*6) (mod 100)
  = 104 (mod 100)
  = 4 (mod 100).
  
  Hence last two digits of 994^499 is 04.
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