Remainder when (2222^5555 + 5555^2222) divided by 7 ?

1) 3
2) 2
3) 1
4) 0

• Sol:
  
  When 5555 and 2222 are divided by 7 we get the remainders 4 and 3 respectively.
  
  so, problem reduces to find the remainder when 42222 + 35555 is divided by 7
  
  so, 42222 + 35555 = (42)1111 + (35)1111
  
  since, an + bn is divisible by (a + b) when n is ODD
  
  therefore, it is divisible by 42 + 35 = 16 + 243 = 259
  
  since, 259 is divisible by 7,
  
  therefore, 7 will divide the 55552222 + 22225555
  
  so, Remainder = 0
  
• 9 years agoHelpfull: Yes(2) No(2)
• 2222= 3 mod 7
  5555 = 4 mod 7
  
  2222^5555 + 5555^2222 = (3^5)^1111 + (4^2)^ 1111
  
  as it is a^n + b^n and n is odd so it is divisible by
  
  3^5 + 4^ 2 = 259
  
  hence by 7 as 7 is a factor of 259
• 8 years agoHelpfull: Yes(0) No(0)