Latest TCS Aptitude Question SOLUTION: 1. P(x)=(x^2012+x^2011+x^2010+…...+x+1)^2-x^2012 Q(x)=x^2011+x^2010+…+x+1 The remainder when P(x) is divided by Q(x) is ?

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1. P(x)=(x^2012+x^2011+x^2010+…...+x+1)^2-x^2012
Q(x)=x^2011+x^2010+…+x+1
The remainder when P(x) is divided by Q(x) is ?

Asked In TCS prashant (11 years ago)

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meena  (11 years ago)

solution is 0.when we submit 1 instead of x we get(2013)^2-1^2012.it is divided by 2011.(by using a^2-b^2 formula)(2014*2012)/2012.remainder is 0.quotient is 2014.

Is this solution Helpfull? Yes (39) | No (4)  

Kabirul Islam  (11 years ago)

remainder will be 2 cause the first one (x^2012+...)^2/(x^2011+.....)=1 and rest portion will also one (like(1/10)remainder 1)
hence 1+1=2

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ankit kumar  (11 years ago)

substitute 1 in place of X..... because if its true for 1 then generalized form of X will be automatically true. why i have put here only 1 is because its the no. that might reduce my calculation time, but you can't take 0 it will led to undefined solution(you may put any no. as you wish).
now for P(0)= 2013^2 - 1 , Q(0)=2012 so remainder will always be 0.

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Srivastav Reddy  (11 years ago)

can u explain clearly..plz.,.,.

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RAVI KIRAN  (11 years ago)

Hi meena your answer seems to be correct, but can u explain why we need substitute 1 as default value , is it an easy way to get the answer? could we go for any other procedure to solve this question

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trupti  (11 years ago)

absolutely correct meena ....very gud....

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Jagjeet   (11 years ago)

answer will be 0.

as p(x)= a^2 - b^2
p(x)= (a-b) (a+b) & (a-b)=Q(x) then the remainder will be 0.

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