Let f(x)= 1+x+x^2+....x^6. what is the remainder when f(x^7) is divided by f(x).

A. 7
B. 6
C. 0
D. none

• geometric progression formula= sum = a( r^m -1)/ (r-1)
  in given series f(x)= 1+x+x^2+x^3+x^4+x^5+x^6
  m=7
  a=1
  r=x
  so, f(x)= x^7-1/(x-1)
  f(x)(x-1)=x^7-1
  f(x)(x-1)+1= x^7
  f[f(x)(x-1)+1]= f(x^7)
  
  f(x^7)mod f(x) = f[f(x)(x-1)+1] mod f(x)
  = f(1)
  = 1+1+1+1+1+1+1= 7 (ans)
• 9 years agoHelpfull: Yes(32) No(2)
• f(x^7)=1+x^7+x^14+x^21.....+x^42
  =1+x^7(1+x^2+x^3+....x^6)
  =1+x^7(f(x))
  since dividend=divisor*quotient+remainder
  hence here remainder is 1
  option D
• 9 years agoHelpfull: Yes(20) No(44)
• Putting x=0 :
  f(x)=1
  f(x^7)=1 Hence, remainder is 0.
  Putting x=1 :
  f(x)=7
  f(x^7)=7 Hence, remainder is 0.
  Keep doing for other nos also, it will come 0 only.
• 9 years agoHelpfull: Yes(11) No(17)
• f(x^7)=1+x7+x14+x21+x28+x35+x42
  =1+X7(1+x7+x14+x21+x28+x35)
  =1+x7(1+x7(1+x7+x14+x21+x28))
  =1+x7(1+x7(1+x7(1+x7+x14+x21)))
  =1+x7(1+x7(1+x7(1+x7(1+x7+x14))))
  =1+x7(1+x7(1+x7(1+x7(1+x7(1+x7))))))
  =1+x7(1+x7(1+x7(1+x7(1+x7(1+x7(1)))))))
  then we can see 1 will always remain till 6 times due to series
  and 1 is alredy present.
  hence total remainder is 6+1=7
• 9 years agoHelpfull: Yes(5) No(0)
• Answer according to TCS seeSame is 7 . (keep trying)
• 9 years agoHelpfull: Yes(3) No(1)
• answer is 1
• 9 years agoHelpfull: Yes(2) No(3)
• putting x=0, f(x)=0
  and f(x^7)=0
  so the remainder =0
• 9 years agoHelpfull: Yes(1) No(8)
• The answer according to them is 7 .
  Someone try !
  
• 9 years agoHelpfull: Yes(0) No(1)
• remainder is 0
• 9 years agoHelpfull: Yes(0) No(3)
• it is coming 7 when i put 2
• 9 years agoHelpfull: Yes(0) No(2)
• not sure.....
  formula: (n-1)!/n=n
• 8 years agoHelpfull: Yes(0) No(1)
• putting x=1
  so f(x)=7
  and f(x^7)=1
  so when 1 is divided by 7 than remainder is 1.
  
• 8 years agoHelpfull: Yes(0) No(0)
• putting x=1
  f(x)=7
  f(x^7)=1
  ans is 7
• 8 years agoHelpfull: Yes(0) No(1)
• ans in campusgate solution(20) last question
• 8 years agoHelpfull: Yes(0) No(0)
• I am unable to understand what is the correct answer... please explain it properly.
  
• 8 years agoHelpfull: Yes(0) No(0)
• Given that f(x7)=1+x7+(x7)2f(x7)=1+x7+(x7)2 + ....+ (x7)6(x7)6 = 1+x7+x14+....+x421+x7+x14+....+x42
  We will rewrite the above equation, f(x7)=1+(x7−1)+(x14−1)+f(x7)=1+(x7−1)+(x14−1)+... + (x42−1)+6(x42−1)+6
  We know that x7−1=(x−1)(x6+x5+...1)x7−1=(x−1)(x6+x5+...1)
  (∵∵ xn−anxn−an = (x−a)(x−a).(xn−1+xn−2.a+(xn−1+xn−2.a+xn−3.a2.....+an−1xn−3.a2.....+an−1 )
  Now It is clear that x7−1x7−1 is exactly divisible by f(x).
  Also x14−1=(x7)2−12x14−1=(x7)2−12 and x7−1x7−1 is a factor of this expression. (∵xn−an∵xn−an is always divisible by x−ax−a
  Similarly, we write x21−1=(x7)3−13x21−1=(x7)3−13, x28−1=(x7)4−14x28−1=(x7)4−14....
  So remainder = 1 + 6 = 7
• 8 years agoHelpfull: Yes(0) No(0)
• 
  12
  iuhni8 btygvu gvuyf
• 7 years agoHelpfull: Yes(0) No(0)
• f(x^7) = 1+x^7+x^14+......x^42
  =1+x^7(1+x^2+......x^6)
  then
  =1+x^7(f(x))
  Dividing by f(x) is 1 mod f(x)=1
• 6 years agoHelpfull: Yes(0) No(0)