What will be the remainder if 4^1000/7

• 4^1000/7
  =(4^3)^333 * 4 / 7
  =(7*9+1)^333*4/7
  =4/7
  => remainder = 4
• 10 years agoHelpfull: Yes(5) No(0)
• whenever we get a form like ,
  (ax+1)^n/a Then , the remainder is always 1
  or
  (ax-1)^n/a Then , the remainder is always (a-1)
  Now ,solving the question by converting the problem in above given forms,
  we get ,
  (((4^3)^333)*4)/7
  Then we are left with ,
  
  1*4/7,
  Which will leave 4 as the remainder.
  Answer=4.
• 10 years agoHelpfull: Yes(2) No(0)
• 4^1000/7
  2^2000/7
  (2^1998*2^2)/7
  ((2^3)^666*2^2)/7
  ((8)^666*2^2)/7
  (1^666*2^2)/7 (since remainder is 1 when 7 divides 8)
  this reduces the problem to 2^2/7
  hence remainder 4
• 10 years agoHelpfull: Yes(1) No(0)
• 4^1000/7
  =(7-3)^1000/7
  =3^1000/7
  =9^500/7
  =2^500/7
  =(((2^3)^166)*4)/7
  =>remainder=4
  
• 10 years agoHelpfull: Yes(1) No(0)
• 4^1000/7
  =(7-3)^1000/7
  =3^1000/7
  =9^500/7
  =2^500/7
  =(((2^3)^166)*4)/7
  =>remainder=4
  
• 10 years agoHelpfull: Yes(1) No(1)
• 4^10^2^2=1.73*10^23
  
• 10 years agoHelpfull: Yes(0) No(0)
• when 4^3/7 rem=1 ,so the expression becomes ((4^3)^333)*4/7 which gives 4 as rem
  
• 10 years agoHelpfull: Yes(0) No(0)
• (remainder of 4^even no=6 and 4^odd no =4.) here 4^1000=6, now 6/7 therefor remainder=4
• 8 years agoHelpfull: Yes(0) No(1)