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

• since,(x^n+y^n) is divisible by (x+y) for odd values of n
  here, x=17,y=29,n=37
  so, 17^37+29^37 is divisible by (17+29)= 46 so divisible by 23 also.
  
  so, remainder is zero.
  
• 10 years agoHelpfull: Yes(7) No(0)
• ans:0
  17^23+29^37/23
  (-6)^37+(6)^37/23
  0/23=23
• 10 years agoHelpfull: Yes(1) No(0)
• since,(x^n+y^n) is divisible by (x+y) for odd values of n
  here, x=17,y=29,n=37
  so, 17^37+29^37 is divisible by (17+29)= 46
  
  so, remainder is zero.
• 10 years agoHelpfull: Yes(1) No(0)
• i think 2^37 as x^n+y^n =(x+y)^n
• 10 years agoHelpfull: Yes(0) No(1)