find the reminder when (2222^5555+5555^2222)/7?

• 0
  
  2222/7=3
  5555/7=4
  
  3^5555+4^2222/7
  (3^5)^1111+(4^2)^1111/7
  3^5/7=5
  4^2/7=2
  5^1111+2^1111/7
  5^1=5
  5^11=*******5
  
  and same for 2^1111
  so last digits after powering both of them must be...5 and 2
  so 5+2/7
  7/7=1
  remainder=0
  
• 10 years agoHelpfull: Yes(1) No(2)