Find out last two digits of (2957^3661) + (3081^3643)

• 98
  using power cycles of 57 and 81
• 11 years agoHelpfull: Yes(22) No(0)
• ans=98, by using cycle of 57 and 81
• 11 years agoHelpfull: Yes(7) No(0)
• Ans is 98
  by solving 57^1 + 81^3
• 11 years agoHelpfull: Yes(7) No(2)
• 57^3661+81^3643
  (57^4)^915*57+41
  (01)915*57+41
  98
• 11 years agoHelpfull: Yes(6) No(0)
• sorry , misprinted on above answer
  the correct way is provided below:
  it is done in this way (7)^1=7
  5*1=5
  so, the number obtain from this is 57.
  similarly (1)^3=1
  8*3=24
  the number obtained is 41.
  now add up 57+41=98....therefore ans= 98
• 11 years agoHelpfull: Yes(5) No(0)
• The answer is 28
• 11 years agoHelpfull: Yes(3) No(9)
• it is done in this way (7)^1=7
  5*1=5
  so, the number obtain from this is 57.
  similarly (1)^3=1
  6*3=18
  the number obtained is 81.
  now add up 57+81=138....therefore ans= 38
• 11 years agoHelpfull: Yes(2) No(5)
• the answer is 60
• 11 years agoHelpfull: Yes(0) No(9)
• 98 using power cycle find last digit of first number as 57 and last digit of second number is 41. now add them the number will be 98. Ans
• 11 years agoHelpfull: Yes(0) No(0)