what is the remainder of 6^17+17^6 is divided by 7?

• 6^17=(7-1)^17
  17^6=(7*2+3)^6
  (7-1)^17+(7*2+3)^6/7
  (-1)^17+3^6/7
  -1+9^3/7
  -1+(7+2)^3/7
  -1+2^3/7
  -1+8/7=0
  
  ans-0
• 12 years agoHelpfull: Yes(15) No(5)
• we can write 6^17 as (-1)^17=-1
  and 17^6 = (14+3)^6 i.e 3^6
  further it can be written as (9)^3= 2^3=8
  
  So, 8-1=7 which is divisible by 7
  Hence answer will be 0 remainder
• 12 years agoHelpfull: Yes(6) No(2)
• 6^17/7= 6
  17^6/7=1
  total remainder=6+1=7
  when divided by 7 again gives remainder 0
  nas will be 0
• 12 years agoHelpfull: Yes(3) No(2)
• split 6^17 into its prime factors as 2^17 * 3^17
  2^17 mod 7 = 4
  3^17 mod 7 = 5
  17^6 mod 7 = 1
  (4*5+1)=21
  21 mod 7 = 0..
  0 is d ans
• 12 years agoHelpfull: Yes(2) No(1)
• remainder is 0
• 12 years agoHelpfull: Yes(1) No(1)
• according to remainder theorem when f(x) is divided by x-a then remainder is f(a)
  
  so write the question in the form of f(6)=6^17 + (2(6)+5)^6
  
  replace 6 by x then f(x)=x^17+(2x+5)^6
  
  write 7 as 6+1
  
  then f(-1) is the answer
• 12 years agoHelpfull: Yes(0) No(0)