Latest TCS Aptitude Question SOLUTION: p(x)=(x^999+x^998+.......x+1)^2-x^999 q(x)=x^998+x^997+.....x+1 the reminder when p(x) is divided by q(x) is a)x+1 b)0 c)1 d)x-1

• Placement Papers
• /
• TCS
• /
• Category

TCS Question Papers: TCS Interview Experience(90) TCS Company Profile

(#M40040486) TCS Category question Keep an EYE

p(x)=(x^999+x^998+.......x+1)^2-x^999
q(x)=x^998+x^997+.....x+1
the reminder when p(x) is divided by q(x) is
a)x+1
b)0
c)1
d)x-1

Asked In TCS anshumala rakesh (9 years ago)

Unsolved Read Solution (10)

Is this Puzzle helpful?   (5)   (3) Submit Your Solution

Read Solution (10) :

srikanth Y  (9 years ago)

lets Assume p(x)=(x^3+x^2+x+1)^2-(x^3)
q(x)=x^2+x+1
if we take x value as 2..p(x)/q(x) Reminder will be zero
ans=0

Is this solution Helpfull? Yes (7) | No   

K Sudharshan reddy  (9 years ago)

substitute x=1
then you get (1000)^2-1/999
i.e., 999999/999 and the remainder is 0

Is this solution Helpfull? Yes (4) | No   

Tamal Chakraborty  (9 years ago)

simply 0

(let x=0 )

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

ABHINAV  (9 years ago)

((x^999+x^998+...+x+1)^2-x^999)/(x^998+x^997+...+x+1)

=> ((x^999+x^998+...+x+1)^2-x^999)/(x^999+x^998+...+x+1)-x^999 or,

=>(x^998+...+x+2)^2-1/(x^998+...+x+2)-1 [taking x^999 as common and cancelling it]

we can now use the identity a^2-b^2=(a+b)*(a-b)

using this we can see that p(x) is clearly divisible by q(x)

i.e. p(x)/q(x)= (x^999+x^998+...+x+1)+x^999

so the remainder will be 0

Is this solution Helpfull? Yes (3) | No   

anudeep  (9 years ago)

Lets assume as x=1,
then p(x)=(1000)^2-1,q(x)=999
p(x)/q(x)=99999/999=0
ans=0.

Is this solution Helpfull? Yes (2) | No   

ramyasree  (9 years ago)

let x be 1 then substitute in p(x) and q(x) now we get 1000/999 then remainder is 1

Is this solution Helpfull? Yes (1) | No (5)  

aditi ojha  (9 years ago)

put x = 1 and solve

eq will be reduced to
(999^2 - 1^2)/998
use identity a^2 - b^2 = (a+b)(a-b)
remainder will be 0

Is this solution Helpfull? Yes (1) | No   

kritee ghosh  (9 years ago)

Anshu,Answer is 0.

Is this solution Helpfull? Yes | No (1)  

priyankajha  (9 years ago)

answer is 0

Is this solution Helpfull? Yes | No   

karthiga  (9 years ago)

p(1)=(1^999+1^998+.......1+1)^2-1^999 = 1000^2 - 1
q(1)=1^998+1^997+.....1+1 = 999
P(1)/Q(1) = 1000^2 - 1^2/999; 1001*999/999 = 1001;
the reminder is 0
A^2 - B^2 = (A+B)* (A-B);

Is this solution Helpfull? Yes | No   

Please Login to Read Solution.