In this question, A^B means A raised to the power B.Let f(x)=1+X+X^2+...X^6.The remainder when f(X^7) is divided by f(X) is:
a)None of the other choices b)6 c)0 d)7

• hey guys it is given in Tcs correct ans is 7
• 10 years agoHelpfull: Yes(13) No(4)
• let x=0 then f(0)=1 and f(0^7)=1 then f(0^7)%f(0)=0 hence c)0 is correct
• 10 years agoHelpfull: Yes(9) No(8)
• Ans is 7
  f(x)=1+x+x^2....x^6
  mathe matics rules
  (x-1)(1+x+x^2...x^6)=x^7-1
  so
  f(x^7)=1+x^7+x^14.....x^42
  f(x^7)%f(x)=1%f(x)+(x^7-1%f(x)+1%f(x)).............because
  (x+1)%y=x%y+1%y
  so
  f(x^7)%f(x)=1+1+1+1+1+1+1=7
  so ans is 7
• 10 years agoHelpfull: Yes(7) No(8)
• option c is the Answer
  f(x)=1+X+X^2+...X^6
  then
  f(x^7)= 1+ x^7 ( f(x)-x ).
  so f(x^7)/f(x) reminder is 0.
  
  by taking x=1 then
  f(x^7)/f(x)= 1+ 1^7(7-1) / 7 =7/7 therefore reminder is 0.
  
  by taking x=2 then
  f(x^7)/f(x)= 1+ 2^7(127-1) / 127
  =1+ 128(126) / 127
  =16129/127
  =127
  
  by taking x=3 then
  f(x^7)/f(x)= 1+ 3^7(1093-1) / 1093
  =1+ 2187(1092) / 1093
  =2388205/1093
  =2185
  
  therefore reminder is 0.
• 10 years agoHelpfull: Yes(6) No(8)
• let say f(x) = y
  f(x^7)= y^7
  (y^7)%(y)=0
• 10 years agoHelpfull: Yes(3) No(13)
• ans 7
  wecant take x=0 as its nt natural nuber
  take x=1
  f(x)=7
  f(x^7)=1
  divide it you will get ans 7
• 10 years agoHelpfull: Yes(2) No(9)
• ans) c.
  f(x)=1+x+...x^6
  let x=1
  f(x)=7
  f(x^7)= 1+x^7+(x^7)^2+...(x^7)^6
  f(x^7)= 7
  remainder= 0
• 10 years agoHelpfull: Yes(2) No(5)
• f(x)=1+x+x^2+...+x^6 put x=1 then
  f(1)=7
  and f(x^7)=7 because at x=1 f(1^7)=1
  so f(x^7)%f(x)=7/7=1 hence remainder will be 0
• 10 years agoHelpfull: Yes(1) No(1)
• f(x)=1+x+x^2+...+x^6 put x=1 then
  f(1)=7
  and f(x^7)=7 because at x=1 f(1^7)=1
  so f(x^7)%f(x)=7/7=1 hence remainder will be 0
• 10 years agoHelpfull: Yes(1) No(3)
• x^0+x^1+......+x^6
  when divided by x^7 7 is prime gives 7
  
• 10 years agoHelpfull: Yes(1) No(0)
• Let x =2 and x^7 = a
  a=128
  f(x)= 1+x^2....... x^6 putting x = 2 in this equation we get f(x) = 127
  a%f(x) = 128%127 = 1
  f(x^7) = 1 + (x^7)^2 + (x^7)^3+........ + (x^7)^6 = 1+a^2 + a^3 + ..... + a^6
  f(x^7)/f(x) = 1/f(x)+a^2/f(x) + a^3/f(x) + ..... + a^6/f(x) = 1+1+1+1+1+1+1 = 7 ans
• 9 years agoHelpfull: Yes(1) No(0)
• Consider f(x) series as GP.
  For f(x), the common ratio is x
  Sum of f(x)'s terms = (x^7 -1)/(x-1)
  Now divide f(x) with each term of f(x^7)
  1*(x-1)/(x^7 - 1) + x^7*(x-1)/(x^7 - 1).......
  Now add +1 and -1 to each term except first and you will get the solution 7.
  
  Suppose x^14/(x^7 - 1)
  Now add +1 and -1,
  ( (x^7)^2 - 1^2 )/ (x^7 -1) + 1 and so on
• 7 years agoHelpfull: Yes(1) No(0)
• let x=2
  f(x)=127
  f(x^7)=128
  thus remainder is 1
  it is valid for all integers
  option a
• 10 years agoHelpfull: Yes(0) No(8)
• 0,because each & every value of x,remainder will be zero.
  
• 10 years agoHelpfull: Yes(0) No(0)
• (x-1)(1+X+X^2+...X^6)=(x^7-1)
  x^7= 1+(x-1)(1+X+X^2+...X^6)
  
  let x^7 = A
  and A%(1+X+X^2+...X^6)=1
  
  f(x^7)=1+x7+x^14+x^21....+x^42
  
  =1+A+A^2+A^3...+A^6
  and when f(x^7)/f(x),the remainder from each term of f(x^7) is 1
  hence 1+1+1+1+1+1+1=7
  remainder is 7.
• 10 years agoHelpfull: Yes(0) No(3)
• 0 is correct ans
  
• 9 years agoHelpfull: Yes(0) No(0)