what would be the remainder when (1111)^2222 + (2222)^3333 is divided by 9?

• (1111)^2222 + (2222)^3333/9
  =(9*123+4)^2222 +(9*247-1)^3333/9
  =4^2222+(-1)^3333/9
  ={(4^3)^740}*4^2 +(-1)/9
  =(9*7+1)^740*16 +(-1)/9
  =16+(-1)/9
  =15/9
  so remainder=6
  
• 11 years agoHelpfull: Yes(21) No(0)
• Ans 6.
  ((1107+4)^2222+(2223-1)^3333)/9
  =(4^2222+(-1)^3333)/9
  =(((4^3)^740)*4^2+(-1))/9
  =((63+1)^740*4^2+(-1))/9
  =(1^740*4^2+(-1))/9
  =(16-1)/9
  rem=6.
• 11 years agoHelpfull: Yes(6) No(2)
• Theorem: a^n+b^n is divisible by a+b when n is odd
  So,
  1111/9 remainder=4
  2222/9 remainder=8
  Hence the problem reduces to finding the remainder when (4^2222+8^3333)/9
  Now using the above stated theorem for (4^2222+8^3333)/9
  We get==== [4^2(1111)+8^3(1111)]
  Again result we get,
  4^2+8^3=528
  If 528 had been a multiple of 9 the remainder would be 0,
  Since it is not so divide 528/9
  The remainder is 6.. ans
  
• 11 years agoHelpfull: Yes(3) No(2)
• ans:2 find 2 digit num
  (1111)^2222=units digit is 1 and tens digit 1*2 so 21
  (2222)^3333=last 2 digits are 44
  total 65 when divided by 9 gives 2
• 11 years agoHelpfull: Yes(1) No(2)
• (4^3)^746*4^2/9=7
  (-1)^3333=-1
  7-1/9=rem 6
  
  
• 11 years agoHelpfull: Yes(0) No(3)