When 2222^5555 + 5555^2222 is divided by 7 ,the remainder is ?

• 2222^5555 / 7
  => (7*317+3)^5555 / 7
  => 3^5555 / 7
  => (3^6)^925 * 3^5 / 7
  => (7*104+1)^925 * (7*34 +5) / 7
  => rem = 5
  
  5555^2222 / 7
  => (7*793 +4)^2222 / 7
  => 4^2222 / 7
  => (4^3)^740 * 3^2 / 7
  => (7*9+1)^740 *(7+2) / 7
  => rem = 2
  
  so, 2222^5555 + 5555^2222 / 7 gives rem (5+2)/7 = > rem = 0
  
  
• 9 years agoHelpfull: Yes(6) No(1)
• In such a question, always remember that :
  
  100/7 leaves remainder 2 so, (100^2)/7 will leave remainder 2*2 = 4. This is the remainder theorem.
  
  Proceeding in this way, let us take the first part i.e., 2222^5555 / 7.
  
  Now, 2222/ 7 =3. Therefore, 2222^5555 / 7 = 3^5555/ 7.
  Now 3^5555 = 3^(5*1111) = 243^1111 (as 3^5= 243).
  Therefore, 3^5555/ 7= 243^1111/ 7. Again, 243 / 7 = 5 and 5^1111 / 7 will still give 5 as remainder.
  Therefore remainder of the first part is 5.
  
  Taking the second part, 5555^2222/ 7, 5555/7= 4. So, 5555^2222/ 7 = 4^2222/7 = 4^(2*1111) / 7
  Similarly as above, 4^ (2*1111) = 16^1111 and 16^1111 / 7 = 2^1111 / 7 ( As 16/7 will give remainder 2).
  And remainder of the second part will be 2^1111/7 which will be 2.
  
  So, sum of the remainders of both sides is 5+ 2= 7. This divided by 7 will give remainder 0 which is the answer.
  
  If you have any queries regarding the method, feel free to ask.
• 9 years agoHelpfull: Yes(5) No(3)
• 13
  since 2 goes in the cycle of 4 and sin the cycle of 1.
  so remainder=8+5=13/7=6 reamainder
• 9 years agoHelpfull: Yes(0) No(0)