Find the number of quadratic polynomials ax² + bx + c such that:
a) a, b, c are distinct.
b) a, b, c ε {1, 2, 3, ...2008}
c) x + 1 divides ax² + bx + c
a) 2013018 b) 2013021 c) 2014024 d) 2018040 e) none of these
Is the answer c)2014024.
ReplyDeleteReasoning->b=a+c
b ranges from 3 to 2008
x1+x2=n has n-1C1 solns.Using th elogic we get
2+3+..2007 solutions. A total of 2015027 solutions.here,condition 1 comes into picture telling us that a,b,c are distinct.Therefore the even values of b ie.4,6,8,..2008 will have one solution less ie.from the given solution subtract 1003 thus giving us 20104024.
PS-Never underestimate a condition :P ..Great question btw.
correct answer
ReplyDeletecan u pls explain it more elaborately
ReplyDeleteExactly which part of my solution are u not able to understand.
ReplyDeleteThe first part
b=a+c is deduced from the fact that x=-1 is a solution of ax^2+bx+c=0.
this implies after substittuing x=-1,we get a-b+c=0. giving us b=a+c
Given a,b ,c are distinct and belong to 1,2,3...2008
therefore
b ranges from 3 to 2008
x1+x2+x3+..xr=n has n-1Cr-1 positive solutions.Using the logic we get
for b=3 has 2C1 ie. 2 solns(a=1,b=2 and a=2 b=1)
similarly b=4,5,6,7...2008 has 3,4,5...2007..ie a total of
2+3+..2007 solutions. A total of 2015027 solutions.here,condition 1 comes into picture telling us that a,b,c are distinct.Therefore the even values of b ie.4,6,8,..2008 will have one solution less ie.from the given solution subtract 1003 thus giving us 20104024.
thnx A lot
ReplyDeleteanother possible solution could be...
ReplyDeleteas c=b-a
when c=1 we will have 2007 pairs((2,1),(3,2)....)
when c=2 we will have 2005 pairs
when c-3 we will have 2005 pairs...
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when c=2007 no. of pair is 1.
hence adding we will get--2104024
I M getting b)
ReplyDeletesince b=a+c
if a=1, c=2,3,4,5,...2007(total solution 2006)
if a=2, c=1,3,4,5....2005(total solution 2005)
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if a=2006, c=1,2(tot sol=2)
if a=2007, c=1(tot sol=1)
adding them all=2006+2005+......+2+1
=2013021
cn any1 tell me where am i wrong..
a+c=b then moving forward:-
ReplyDeleteFor a=1(let us assume c>a),c could take 2006 values such that a,b,c are all different.For example:-
a=1,c=2,b=3
a=1,c=3,b=4
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a=1,c=2007,b=2008
Similarly for a=2,c could take 2004 values.
for a=3,c could take 2002 values.
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for a=1003,c could take 2 values.
Now no. of solutions is 2+4+6+8....+2006=1014012
c<a cases are also there.Therefore ans=2*1014012
i am not satisfied with the abhitsian sol. because ,as he has mention in for b=4we have 3distinct set of values of (a,c) i.e is (1,3) (3,1), (2,2) which simply contradicts the basic conditin of the question
ReplyDelete@ayashgupta towards the end abhitsian is subtracting all those cases where u do not have distinct values for a,b and c .He is no where contradicting the basic condition of the question as u say
ReplyDelete