p(x)=(x^999+x^998+.......x+1)^2-x^999
q(x)=x^998+x^997+.....x+1
the reminder when p(x) is divided by q(x) is
a)x+1
b)0
c)1
d)x-1

• lets Assume p(x)=(x^3+x^2+x+1)^2-(x^3)
  q(x)=x^2+x+1
  if we take x value as 2..p(x)/q(x) Reminder will be zero
  ans=0
• 10 years agoHelpfull: Yes(7) No(0)
• substitute x=1
  then you get (1000)^2-1/999
  i.e., 999999/999 and the remainder is 0
• 10 years agoHelpfull: Yes(4) No(0)
• simply 0
  
  (let x=0 )
• 10 years agoHelpfull: Yes(4) No(4)
• ((x^999+x^998+...+x+1)^2-x^999)/(x^998+x^997+...+x+1)
  
  => ((x^999+x^998+...+x+1)^2-x^999)/(x^999+x^998+...+x+1)-x^999 or,
  
  =>(x^998+...+x+2)^2-1/(x^998+...+x+2)-1 [taking x^999 as common and cancelling it]
  
  we can now use the identity a^2-b^2=(a+b)*(a-b)
  
  using this we can see that p(x) is clearly divisible by q(x)
  
  i.e. p(x)/q(x)= (x^999+x^998+...+x+1)+x^999
  
  so the remainder will be 0
• 10 years agoHelpfull: Yes(3) No(0)
• Lets assume as x=1,
  then p(x)=(1000)^2-1,q(x)=999
  p(x)/q(x)=99999/999=0
  ans=0.
• 10 years agoHelpfull: Yes(2) No(0)
• let x be 1 then substitute in p(x) and q(x) now we get 1000/999 then remainder is 1
• 10 years agoHelpfull: Yes(1) No(5)
• put x = 1 and solve
  
  eq will be reduced to
  (999^2 - 1^2)/998
  use identity a^2 - b^2 = (a+b)(a-b)
  remainder will be 0
• 10 years agoHelpfull: Yes(1) No(0)
• Anshu,Answer is 0.
• 10 years agoHelpfull: Yes(0) No(1)
• answer is 0
  
• 10 years agoHelpfull: Yes(0) No(0)
• p(1)=(1^999+1^998+.......1+1)^2-1^999 = 1000^2 - 1
  q(1)=1^998+1^997+.....1+1 = 999
  P(1)/Q(1) = 1000^2 - 1^2/999; 1001*999/999 = 1001;
  the reminder is 0
  A^2 - B^2 = (A+B)* (A-B);
  
• 10 years agoHelpfull: Yes(0) No(0)