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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
Read Solution (Total 15)
-
- 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
- 9 years agoHelpfull: Yes(49) No(46)
- 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.
- 9 years agoHelpfull: Yes(36) No(3)
- 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 - 9 years agoHelpfull: Yes(31) No(0)
- 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.
- 9 years agoHelpfull: Yes(7) No(0)
- for x= 0, remainder is 0
for x= 1, remainder is 0
for x= -1, remainder is 0 - 9 years agoHelpfull: Yes(4) No(5)
- 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 - 9 years agoHelpfull: Yes(4) No(2)
- rohit mohan srivastava is correct.rakesh is wrong
- 9 years agoHelpfull: Yes(3) No(0)
- answer is: 7
- 9 years agoHelpfull: Yes(2) No(2)
- 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
- 9 years agoHelpfull: Yes(2) No(5)
- 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... - 9 years agoHelpfull: Yes(1) No(4)
- its answer is not 0
- 9 years agoHelpfull: Yes(0) No(2)
- 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
- 9 years agoHelpfull: Yes(0) No(1)
- 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
- 8 years agoHelpfull: Yes(0) No(0)
- 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. - 6 years agoHelpfull: Yes(0) No(0)
- 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 - 5 years agoHelpfull: Yes(0) No(0)
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