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.

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.

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

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..

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 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

RT @pbhushan1: Kejriwal screaming vendetta merely because his 'trusted man' is raided by CBI w/o informing is absurd.Should an accused be w… 2 months ago

RT @rahulkanwal: The allegations against Rajender Kumar were levelled by Ashish Joshi, member Secy, Delhi dialogue Commission. CBI acting o… 2 months ago

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.

abhitsianJuly 27, 2009 at 9:19 am

correct answer

RahulJuly 27, 2009 at 8:31 pm

can u pls explain it more elaborately

saurabhJuly 28, 2009 at 10:44 am

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.

abhitsianJuly 28, 2009 at 12:36 pm

thnx A lot

saurabhJuly 28, 2009 at 5:10 pm

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 PANGHALAugust 10, 2009 at 10:40 am

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..

SetuAugust 13, 2009 at 3:24 pm

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 DudejaAugust 18, 2009 at 2:34 pm

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

ayashguptaNovember 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

BharatAugust 29, 2010 at 3:35 pm