Latest TCS Aptitude Question SOLUTION: P ( x ) = ( x2012 + x2011 + x2010 + . . . . + x + 1)2 – x2012 Q ( x ) = x2011 + x 2010 + . . . . + x + 1 The reminder when P ( x ) is divided by Q ( x ) is

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(#M40024299) TCS Numerical Ability question Keep an EYE

P ( x ) = ( x2012 + x2011 + x2010 + . . . . + x + 1)2 – x2012
Q ( x ) = x2011 + x 2010 + . . . . + x + 1
The reminder when P ( x ) is divided by Q ( x ) is

Asked In TCS suryakumar jaidi (10 years ago)

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Amrit Chowdhury  (10 years ago)

Answer:-remainder is zero.

substitute 1 in x
then it will become
((2013)^2-1)/2012
(2013+1)(2013-1)/2012
2014 is quotient
So, remainder is '0'.

Is this solution Helpfull? Yes (19) | No (2)  

Amrit Chowdhury  (10 years ago)

Answer:-remainder is zero.

P(x)=(x^2012+x^2011+x^2010+...+x^2+x+1)^2-x^2012
substitute x=1
P(x)=(1+1+1+.......2013 terms)^2-(1)^2012 =2013^2-1^2
Q(x)= x^2011 + x^2010 + . . . . + x + 1
again substitute x=1
Q(x)=1+1+.....+2012 terms =2012

So,P(x)/Q(x)=
((2013)^2-1)/2012 =((2013)^2-(1)^2)/2012
(2013+1)(2013-1)/2012[From Formula(a^2-b^2)=(a+b)(a-b)]
=(2014*2012)/2012
2014 is quotient
So, remainder is '0'.

Is this solution Helpfull? Yes (10) | No (1)  

jaya akula  (10 years ago)

amrit can u plz explain y did 2014 became quotient...as well as how 0 is the remainder..plz xplain in detail..

Is this solution Helpfull? Yes (3) | No (2)  

Amrit Chowdhury  (10 years ago)

(x^2012+x^2011+x^2010+...+x^2+x+1)^2 is given in question so if we put x=1,there are total 2013 terms so 2013^2..

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ipsita saha  (10 years ago)

plz explain how is it 2013^2

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Himanshu tamrakar  (10 years ago)

(x2012/x2011+....+x+1)+( x2011 + x2010 + . . . . + x + 1)/(x2011 + x 2010 + . . . . + x + 1)than reminder will x2012

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Abhijit Das  (10 years ago)

can we put 0 in both the func ,so p(0) =1 as well as q(0) = 1 ,so remainder is 0 ??
plss help

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jaya akula  (10 years ago)

amrit can u tell me the division part...as quotient is 2014..what are the values of divisor and dividend...??

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