Latest TCS Aptitude Question SOLUTION: a^b means a raised to the power of b .let f(x)=1+x+x^2+........x^6.the remainder when f(x^7) is divided by f(x) is

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(#M40039000) TCS Numerical Ability question Keep an EYE

a^b means a raised to the power of b .let f(x)=1+x+x^2+........x^6.the remainder when f(x^7) is divided by f(x) is

Asked In TCS jass (9 years ago)

Unsolved Read Solution (15)

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Read Solution (15) :

RAKESH  (9 years ago)

to find remainder of a polynomial fn f(x)/g(x)
put x = 0,1 or any value

put x=0
f(x)=1
f(x^7)=1
f(x^7)/f(x)= 1/1 => remainder = 0

put x=1
f(x)=7
f(x^7)=7
f(x^7)/f(x)= 7/7 => remainder = 0

Is this solution Helpfull? Yes (49) | No (46)  

Rohit Shaw  (9 years ago)

There is a pattern.
if , f(x) = 1+x+x^2+.....+x^(n).
for x > 1,
then the remainder when f(x^(n+1)) is divided by f(x) is (n+1).
so , here remainder is 7.

Is this solution Helpfull? Yes (36) | No (3)  

ROHIT MOHAN SRIVASTAVA  (9 years ago)

answer is 7..put x=2
f(x)=127
f(x^7)=1+128+128^2+128^3+128^4+128^5+128^6
f(x^7)/f(x)=1/127+128/127+128^2/127+128^3/127+128^4/127+128^5/127+128^6/127
=1+1+1+1+1+1+1=7

Is this solution Helpfull? Yes (31) | No   

shubhankar  (9 years ago)

In this question there should be a condition which is x>1, for x=2,3... the remainder is 7. so correct ans is 7.

Is this solution Helpfull? Yes (7) | No   

MUNISH YADAV  (9 years ago)

for x= 0, remainder is 0
for x= 1, remainder is 0
for x= -1, remainder is 0

Is this solution Helpfull? Yes (4) | No (5)  

Sruthi Keerthi Bandi  (9 years ago)

f(x)=1+x+x^2+x^3+....+x^6 = 1*(x^7-1)/6. /*sum of n terms in gp = a(r^n-1)/(r-1)*/
f(x^7)=1+x^7+(x^7)^2+(x^7)^3+...+(x^7)^6 = 1*((x^7)^2-1)/6.
f(x^7)/f(x) = x^7+1

Is this solution Helpfull? Yes (4) | No (2)  

abhisek banerjee  (9 years ago)

rohit mohan srivastava is correct.rakesh is wrong

Is this solution Helpfull? Yes (3) | No   

sneha  (9 years ago)

answer is: 7

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

aman sharma  (9 years ago)

f(x^7)= 1+x^7+x^14......x^42
we can write it as = 1+x^7(1+x^2+.....x^6)
( 1+x^7(f(x)) ) /f(x)
then rem=1

Is this solution Helpfull? Yes (2) | No (5)  

Preethi Mohan  (9 years ago)

if we substitute 0, in place of X,

say f(0^7)/f(0)=1/1.

remainder=0,
if we substitute 1 in place of x
f(1^7)/f(1) = 7/7 = 0

but from 2,3,4....... the remainder will be 1

let me give an example substitute 2,

f(2) = 1+2+4+8+16+32+64 = 127 and
f(3) = 1+3+9+27+81+243+729 = 1093

f(2^7)/f(2) = 1+ (multiple of 127)/127, Because x+x^2+x^3.....all are multiple of x

remainder will be 1

likewise

f(3^7)/f(3)=1+(multiple of 1093)/1093

so the remainder will be 1....

they must have given the conditions where x>1 or something else to predict the correct result...

Is this solution Helpfull? Yes (1) | No (4)  

purva  (9 years ago)

its answer is not 0

Is this solution Helpfull? Yes | No (2)  

md abdulla choush  (9 years ago)

to find remainder of a polynomial
fn f(x)/g(x)
put x = 0,1 or any value

put x=0
f(x)=1
f(x^7)=1
f(x^7)/f(x)= 1/1 => remainder = 0

put x=1
f(x)=7
f(x^7)=7
f(x^7)/f(x)= 7/7 => remainder = 0

Is this solution Helpfull? Yes | No (1)  

neeraj  (8 years ago)

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

Is this solution Helpfull? Yes | No   

Anup Kumar Singh  (6 years ago)

f(x)=1+x+x^2......+x^6.
sum of the G.P series = (x^7-1)/(x-1).
put this "(x^7-1)/(x-1)"=0;
Get value of x=1.
put this value of x=1 in f(x^7)=1+x^7+.............+(x^7)^6. You we get the remainder=7.

Is this solution Helpfull? Yes | No   

Prerna bisen  (5 years ago)

Given that
f
(
x
7
)
=
1
+
x
7
+
(
x
7
)
2
f(x7)=1+x7+(x7)2
+ ....+
(
x
7
)
6
(x7)6
=
1
+
x
7
+
x
14
+
.
.
.
.
+
x
42
1+x7+x14+....+x42

We will rewrite the above equation,
f
(
x
7
)
=
1
+
(
x
7
−
1
)
+
(
x
14
−
1
)
+
f(x7)=1+(x7−1)+(x14−1)+
... +
(
x
42
−
1
)
+
6
(x42−1)+6

We know that
x
7
−
1
=
(
x
−
1
)
(
x
6
+
x
5
+
.
.
.1
)
x7−1=(x−1)(x6+x5+...1)

(
∵
∵
x
n
−
a
n
xn−an
=
(
x
−
a
)
(x−a)
.
(
x
n
−
1
+
x
n
−
2
.
a
+
(xn−1+xn−2.a+
x
n
−
3
.
a
2
.
.
.
.
.
+
a
n
−
1
xn−3.a2.....+an−1
)
Now It is clear that
x
7
−
1
x7−1
is exactly divisible by f(x).
Also
x
14
−
1
=
(
x
7
)
2
−
1
2
x14−1=(x7)2−12
and
x
7
−
1
x7−1
is a factor of this expression. (
∵
x
n
−
a
n
∵xn−an
is always divisible by
x
−
a
x−a

Similarly, we write
x
21
−
1
=
(
x
7
)
3
−
1
3
x21−1=(x7)3−13
,
x
28
−
1
=
(
x
7
)
4
−
1
4
x28−1=(x7)4−14
....
So remainder = 1 + 6 = 7

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