If X^Y denotes X raised to the power of Y, find out last two digits of
(2957^3661) + (3081^3643)

• (2957^3661) + (3081^3643)
  7^1=7 1^3=1
  5*1=5 8*3=24
  last 2 digits 57 last 2 digits 41
  den, 57+41=98..... ans 98
• 9 years agoHelpfull: Yes(6) No(0)
• ans is 57+41=98
• 9 years agoHelpfull: Yes(3) No(1)
• 7^3, the last digit becomes 1. so that write (2957)3661. 3661 in the form cubes
  3661/3=1220
  1220/3=406 and another 2 is remaining
  therfore (2957)61220 .(2957)^3=1
  (3081)3641=1
  1+1=2
  therfore the last term becomes 2
  
• 9 years agoHelpfull: Yes(2) No(0)
• 2957^3661 THE LAST TWO DIGIT WILL BE 07 BECAUSE 07 IS REPEATED AFTER 4 TIMES
  AND (3081)^3643 THE LAST TWO DIGIT WILL BE 41
  SO ANS 07+41
• 9 years agoHelpfull: Yes(0) No(3)
• 57^1 + 81^3=57+41=98
• 9 years agoHelpfull: Yes(0) No(0)