Latest TCS Aptitude Question SOLUTION: 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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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 M.R.kartheek (11 years ago)

Solved Atul Read Solution (6)

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Read Solution (6) :

Amit kumar  (11 years ago)

Put x=1
P(1)= 2013^2-1
Q(1)=2012 = 2013-1
P(1) divided by Q(1) is remainder 0.

Is this solution Helpfull? Yes (17) | No (2)  

Atul  (11 years ago)

P(x) =[x^2012 + Q(x)]^2-x^2012

P(x) =x^4024+Q^2+2*Q*x^2012 -x^2012
P(x) =Q^2+2*Q*x^2012+(x^2012)*(x^2012-1)

Now x^2012-1 = Q*(x-1)

P(x) = Q^2 +2Qx^2012 +x^2012*Q*(x-1)

As all terms of P are having Q as a factor,
so P is fully divisible bt Q.

so zero remainder.

Is this solution Helpfull? Yes (6) | No (1)  

Sushma  (11 years ago)

@ Karteek,

What is the fun in copy pasting the puzzle and solution yourself?

I do not think admin is giving marks in such cases.

Enjoy and have fun.:-)

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

bharat solanke  (11 years ago)

p(x)=x^2012+x^2011+....+x+1-x^2012
so x^2012 must be cut (x^2012-x^2012)=0
p(x)=x^2011+....+x+1
q(x)=x^2011+...+x+1

Is this solution Helpfull? Yes (3) | No   

M.R.kartheek  (11 years ago)

P(x)= x^2012+x^2011+x^2010+...+x+1)^2 - x^2012
Q(x)= x^2011+x^2010+..+x+1
Q(x) = sum of 2012 terms of G.P
= 1*(x^2012 - 1)/(x - 1)
so
P(x)= (x*Q(x)+1)^2 - x^2012
then
P(x)/Q(x) = (x*Q(x)+1)^2 - x^2012 / Q(x)

(x*Q(x)+1)^2 can be written as:
(x*Q(x))^2 + 2*x*Q(x) + 1

P(x)/Q(x) = ((x*Q(x))^2 + 2*x*Q(x) + 1 - x^2012) / Q(x)
so finally take a look :
P(x)/Q(x) = (1 - x^2012) / Q(x)

P(x)/Q(x) = (1 - x^2012) / (x^2012 - 1) / (x - 1)

so clearly is getting P(x)/Q(x) = (1-x)
enjoy.....

Is this solution Helpfull? Yes (1) | No (9)  

Atul  (11 years ago)

Remainder will be zero.

Let others try.

I shall give details later.

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

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