what is the remainder when 6^17+117^6 is divided by 7?

• 6/7 remainder = 6
  6^2/7 = 36/7 remainder = 1
  6^3/7 = 216/7 remainder = 6
  similarly 6^17/7 remainder = 6
  comes to 117^6 in number theory there is fermat's theorem
  a^(p-1) = 1mod p here p = 7 a 117 so remainder will be 1
  now above problem can be written as 6^17/7 + 117^6/7
  or remainder 6/7+1/7
  or 7/7
  so 0 will be the remainder
• 12 years agoHelpfull: Yes(12) No(0)
• as 6/7=remainder =-1
  117/7=remainder=-2
  so use remainder theorem
  (-1)^17+(-2)^6/7
  64-1/7
  63/7
  remainder=0
• 12 years agoHelpfull: Yes(6) No(1)
• 6^17+117^6=(7-1)^17+(119-2)^6
  after breaking it by binomial theorem we get the last term from
  each one as:-17C17*(-1)^17 = -1 and another one 6C6*(-2)^6 = 64.
  so sum is 63 and divided by 7 so reminder:-0.
  
• 12 years agoHelpfull: Yes(5) No(0)
• remainder=0
  
• 12 years agoHelpfull: Yes(3) No(1)
• unit digit of in power for 6^17 is 7.
  unit digit of in no for 117^6 is 7.
  so 7/7 then remainder is zero.
  
• 12 years agoHelpfull: Yes(1) No(0)
• (6*6^16+117^6)/6=(6(1^8)+1^6)/7=7/7,so remainder 0.
  
• 9 years agoHelpfull: Yes(0) No(0)
• (6^17/7)+(117^6)/7
  (-1)17+(5^6)/7
  -1+((7-2)^6)/7
  -1+2^6/7
  -1+((2^3)62)/7
  -1+1
  0
  so remainder is 0
• 9 years agoHelpfull: Yes(0) No(0)
• 6^17 mod 7=(7-1)^17 mod 7=(-1)^17 mod 7=-1
  117^6mod 7=(7*16+5)^17mod7=(5)^17 mod7=1
  the remainder when 6^17+117^6 is divided by 7 is
  1-1=0(ans)
• 6 years agoHelpfull: Yes(0) No(0)