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Problem of the day 27.07.09

with 10 comments


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

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Written by Implex

July 27, 2009 at 7:50 am

10 Responses

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  1. Is the answer c)2014024.

    Reasoning->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.

    abhitsian

    July 27, 2009 at 9:19 am

  2. can u pls explain it more elaborately

    saurabh

    July 28, 2009 at 10:44 am

  3. Exactly which part of my solution are u not able to understand.
    The 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.

    abhitsian

    July 28, 2009 at 12:36 pm

  4. another possible solution could be…
    as 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…
    .
    .
    .
    when c=2007 no. of pair is 1.
    hence adding we will get–2104024

    ANKIT PANGHAL

    August 10, 2009 at 10:40 am

  5. I M getting b)
    since 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)
    .
    .
    .
    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..

    Setu

    August 13, 2009 at 3:24 pm

  6. a+c=b then moving forward:-

    For 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
    .
    .
    .
    .
    a=1,c=2007,b=2008

    Similarly for a=2,c could take 2004 values.
    for a=3,c could take 2002 values.
    .
    .
    .
    .
    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

    A nkur Dudeja

    August 18, 2009 at 2:34 pm

  7. 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

    ayashgupta

    November 24, 2009 at 1:49 am

    • @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

      Bharat

      August 29, 2010 at 3:35 pm


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