The remainder when 17^37+29^37 is divided by 23 is

• Remainder is 0
  
  (17)^x + (29)^x is always divisible by 23 where x is an odd number
  
  Eg:- 17^3 + 29^3 divisible by 23
  
  => (17)^37 + (29)^37 divisible by 23
  => remainder is 0.
• 11 years agoHelpfull: Yes(2) No(1)
• If n is odd, a^n + b^n = (a + b)(a^(n - 1) - a^(n - 2)*b + a^(n - 3)*b^2 -... + b^(n - 1)).
  Therefore, a^n + b^n is divisible by (a + b) when n is odd
  
  so when 17^37+29^37 is divided by 23, remainder is zero as (17^37+29^37) is divisible by (17+29) or 46 which is divisible by 23.
• 11 years agoHelpfull: Yes(2) No(0)
• when we apply a remainder theorem then we will get (-6)^37 + (6)^37 and the power is 37 it is odd so (-6)^37 is negative value and
  (6)^37 is positive value, both magnitudes are equal then the remainder is zero...
• 11 years agoHelpfull: Yes(1) No(0)
• 0
  37(17+29)/23=74
  remainder 0
  
• 11 years agoHelpfull: Yes(0) No(2)