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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 ?
Read Solution (Total 6)
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- Put x=1
P(1)= 2013^2-1
Q(1)=2012 = 2013-1
P(1) divided by Q(1) is remainder 0. - 12 years agoHelpfull: Yes(17) No(2)
- 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. - 12 years agoHelpfull: Yes(6) No(1)
- @ 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.:-) - 12 years agoHelpfull: Yes(4) No(1)
- 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 - 12 years agoHelpfull: Yes(3) No(0)
- 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..... - 12 years agoHelpfull: Yes(1) No(9)
- Remainder will be zero.
Let others try.
I shall give details later. - 12 years agoHelpfull: Yes(1) No(3)
TCS Other Question
A bag contains printed articles of 4 different kinds: periodicals, novels, newspapers
and hardcovers. When 4 articles are drawn from the bag without replacement, the
following events are equally likely:
the selection of 4 periodicals
the selection of 1 novel and 3 periodicals
the selection of 1 newspaper, 1 novel and 2 periodicals and
the selection of 1 article of each kind
What is the smallest number of articles in the bag satisfying these conditions? How
many of these are of each kind?
1. Alok and Bhanu play the following game on arithmetic expressions. Given the
expression
N = (Θ + A)/B + (Θ + C + D)/E
where A, B, C, D and E are variables representing digits (0 to 9), Alok would like to
maximize N while Bhanu would like to minimize it. Towards this end, they take turns in
instantiating the variables. Alok starts and, at each move, proposes a value (digit 0-9)
and Bhanu substitutes the value for a variable of her choice. Assuming both play to their
optimal strategies, what is the value of N at the end of the game? Also find a sequence
of moves (digits by Alok and variables by Bhanu) that would yield this value.
Note: Moves that lead to a divide-by-zero condition are disallowed. A non-optimal
sequence of moves is (5 → B, 6 → C , 3 → D, 2 → E, 0 → A) and the expression
evaluates to Θ/5 + (Θ+9)/2.
2. The mean, unique mode, median and range of 21 positive integers is 21. What is the
largest value that can be in this sequence? Also find such a sequence.
Note: Given a sequence of numbers a(1) ≤ a(2) ≤ ... ≤ a(n),
The median of the sequence is the middlemost value in the sequence if n is
odd and the average of the two middle values if n is even.
The mode is the most occurring value in the sequence
The range is the difference between the largest and the smallest values, i.e.
a(n) - a(1).
For example, the sequence 2, 3, 4, 6, 6, 9 has mean = (2 + 3 + 4 + 6 + 6 + 9)/6 = 5,
median = (4+6)/2 =5, mode = 6, and range = 9 – 2 = 7.