Latest TCS Aptitude Question SOLUTION: 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

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(#M40038835) TCS Numerical Ability question Keep an EYE

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

Asked In TCS Varsha (9 years ago)

Unsolved Read Solution (18)

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Read Solution (18) :

Niveditha Ravichandran  (9 years ago)

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)

Is this solution Helpfull? Yes (32) | No (2)  

Nirmal Kumar Acharya  (9 years ago)

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

Is this solution Helpfull? Yes (20) | No (44)  

MUNISH YADAV  (9 years ago)

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.

Is this solution Helpfull? Yes (11) | No (17)  

prasoon saxena  (9 years ago)

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

Is this solution Helpfull? Yes (5) | No   

Varsha  (9 years ago)

Answer according to TCS seeSame is 7 . (keep trying)

Is this solution Helpfull? Yes (3) | No (1)  

payal yadav  (9 years ago)

answer is 1

Is this solution Helpfull? Yes (2) | No (3)  

ESHA  (9 years ago)

putting x=0, f(x)=0
and f(x^7)=0
so the remainder =0

Is this solution Helpfull? Yes (1) | No (8)  

Varsha  (9 years ago)

The answer according to them is 7 .
Someone try !

Is this solution Helpfull? Yes | No (1)  

Deepak kumar  (9 years ago)

remainder is 0

Is this solution Helpfull? Yes | No (3)  

swap  (9 years ago)

it is coming 7 when i put 2

Is this solution Helpfull? Yes | No (2)  

shiva kumar  (8 years ago)

not sure.....
formula: (n-1)!/n=n

Is this solution Helpfull? Yes | No (1)  

preeti  (8 years ago)

putting x=1
so f(x)=7
and f(x^7)=1
so when 1 is divided by 7 than remainder is 1.

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deepika singh  (8 years ago)

putting x=1
f(x)=7
f(x^7)=1
ans is 7

Is this solution Helpfull? Yes | No (1)  

jatin bajpai  (8 years ago)

ans in campusgate solution(20) last question

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Akash Shakya  (8 years ago)

I am unable to understand what is the correct answer... please explain it properly.

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Rajat  (8 years ago)

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

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Suryadeep Kundu  (7 years ago)

12
iuhni8 btygvu gvuyf

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SOORIYA  (6 years ago)

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

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