Number of Quadratic equation with different condition

Question

If the number of quadratic polynomials $ax^2+2bx+c$ which satisfy the following conditions:

(i) a, b, c are distinct

(ii) a, b, c $\in$ ${1, 2, 3,. 2001, 2002}$

(iii) x + 1 divides $ax^2+2bx+c$

is equal to $1000 \lambda$, then find the value of $\lambda$.

My solution is as follow

x + 1 divides hence $f(x)=ax^2+2bx+c$, hence $f(-1)=0$, therefore $2b=a+c$

As a.b.c are distinct therefore $a\ne c$

Maximum value of b is 1000 and minimum value of b is 1 and all the value increases by 1

Minimum value of a+c=2 and Maximum value of a+c=2000

It is like distribution of 2b identical balls into 2 distinct boxes so that each box has at least 1 balls ${}^{2b - 1}{C_{2 - 1}} \Rightarrow {}^{2b - 1}{C_1}$

b=1; ${}^{2 - 1}{C_{2 - 1}} = 1$ b=2; ${}^{4 - 1}{C_{2 - 1}} = 3$

b=1000, ${}^{2000 - 1}{C_{2 - 1}} = 1999$ The total number of ways=$\frac{n}{2}\left( {2a' + \left( {n - 1} \right)d'} \right) = \frac{{1000}}{2}\left( {2 + \left( {1000 - 1} \right)2} \right) = {1000^2}$

Now as a,b and c are distinct removing the following cases {1,1},{2,2},....{1000,1000} Hence total number of cases are ${\left( {1000} \right)^2} - 1000 = 1000 \times 999, \lambda = 999$ but answer is $2002$, i cannot figure out my mistake

Answer 1

Let us understand the condition $a+c=2b$, where $a,b,c$ are natural numbers. Here $2b$ is even, which can be obtained only by adding two numbers of same parity. Among first $2002$ natural numbers, there are $1001$ odd and $1001$ even numbers. Hence number of ordered pairs $(a,c)$ should be given by $$2\times \left( \binom{1001}{2}+\binom{1001}{2} \right) = 1000 \times 2002$$

Answer 2

(This problem actually has very little to do with quadratic polynomials...)

I guess I've been looking at a lot of arithmetic progression problems lately, so I'm currently primed to think of them when I see $ \ a + c \ = \ 2b \ \ . $ Since $ \ a \ , \ b \ , \ c \ $ are to be distinct integers, the progressions must have non-zero integer differences between their terms. Also, as MyMolecules observes, the parity of $ \ a \ $ and $ \ c \ $ must be the same.

If we start with ascending progressions, those with initial term $ \ 1 \ $ are $ \ 1 , 2 , 3 \ \ , \ \ 1 , 3 , 5 \ \ , \ \ldots \ , $ $ 1 , 1001 , 2001 \ \ , \ \ $ which number $ \ 3 + (n - 1)·2 \ = \ 2001 \ \Rightarrow \ n \ = \ 1000 \ \ . $ Those with initial term $ \ 3 \ \ -- \ \ 3 , 4 , 5 \ \ , \ \ 3 , 5 , 7 \ \ , \ \ldots \ , \ \ 3 , 1002 , 2001 \ \ -- \ $ number $ \ 999 \ \ , $ and so forth up to $ \ 1999 , 2000 , 2001 \ \ . $ So there are $ \ 1000 + 999 + 998 + \ \ldots \ + 1 \ = \ \frac{1000·1001}{2} \ \ $ of these. The ascending progressions with even initial terms are $$ 2 , 3 , 4 \ \ , \ \ 2 , 4 , 6 \ \ , \ \ldots \ , \ \ 2 , 1002 , 2002 \ \ , \ \ 4 , 5 , 6 , \ \ldots \ , \ \ 4 , 1003 , 2002 , \ \ldots \ , \ \ 2000 , 2001 , 2002 \ \ . $$ There are also $ \ \frac{1000·1001}{2} \ $ of these.

We form descending progressions by reversing the order of $ \ a \ $ and $ \ c \ \ , $ so the total number of these is the same as the number of ascending progressions. Hence, the number of sets of quadratic coefficients satisfying the given conditions is $ \ 2 · 2 · \frac{1000·1001}{2} \ = \ 2002·1000 \ \ , $ or $ \ \mathbf{\lambda \ = \ 2002 } \ \ . $