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Find the number of quadratic polynomials ax²+bx+c ,which satisfy the following conditions : (I) a,b,c are distinct (II) a,b,c ∈ {1,2,3,4......999} (III) (x+1) divides (ax²+bx+c)

Solution by me: Since all of coefficients lie between 999 and 1 I wrote a≤999 b≤999 c≤999 now since ax²+bx+c=0 has a root -1 since it is divisible by x+1 so we have a(-1)²+b(-1)+c=0 a-b+c=0 a+c=b≤999 . Hence I tried find the number of solutions for which a+c=999 a+c=998...etc. under the given conditions. First we are going to look at the odd values of b a+c=999 let's take the middle solution that is 499+500=999 and then decrease the numbers individually

499+500=999 498+501=999 497+502=999 up to 3+996=999 2+997=999 and lastly 1+998=999 By looking the first number that's being added in the equations we have 499 ,498.. until 1 .there 499 numbers from 1 to 499 hence we have 499 distinction solutions for a,c such that b=999 and the solutions a+c=b and c+a=b considered the same. However we consider them different as they lead to different quadratic equations hence there are twice as many solutions. which 499(2)=998 Similarly we look at 997=498+499 which has 498(2)=996 solutions and I the end we have a+c=5 which has 2 solutions c+a=3 has 1 solution and a+c=1 which has no solutions since 0 not allowed. In the end we have 2(499+498+497......1) which by the sum of natural number formula is 249500

Similarly for even numbers but for even number there is case where a=c which is not allowed like a+c=998 has 499+499=998 as solution so we shall start looking from 500+498=998 to get 498 solutions Similarly for the even numbers we have 2(498+497+496...3+2+1)=248502. Adding the 2 we have 498002. However the answer is 1996002 please explain the mistake

Vignesh
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1 Answers1

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You are right by iterating on $b$ after finding out $a + c = b$.

$b = 999, 998, \cdots ,3$.

Now, if $b = 999$, add the number of positive solutions to $x_1 + x_2 = 999$. Note that since $999$ is odd, $x_1 \neq x_2$. So, add $\binom{999 - 1}{2 - 1} = 998$.

Similarly, if $b = 998$, add $\binom{998 - 1}{2 - 1} = 997$, but subtract $1$ because $x_1 = x_2 = 499$ is possible. So, add $996$.

A.S.O.

Ans = $$998 + (997 - 1) + 996 + (995 - 1) + \cdots + (3 - 1) + 2$$ $$= 2 + 3 + \cdots + 998 - 498$$ $$=\frac{998 \cdot 999}{2} - 499$$ $$=499 \cdot 998$$ $$=498,002$$

MangoPizza
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  • Yes very thanks I got the same but the answer given in the book is 196002 – Vignesh Apr 17 '22 at 06:14
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    +1: In my opinion, your work is valid and your answer is accurate. – user2661923 Apr 17 '22 at 06:26
  • @Vignesh Please re-check the constraints of the problem. – user2661923 Apr 17 '22 at 06:42
  • @user2661923 Hello I didn't actually understand what you meant. Do you mean that the answer got by me and mangopizza(498002) right and that the problem is with the question? – Vignesh Apr 17 '22 at 07:47
  • @Vignesh Yes, that is exactly what I mean. Besides the inconsistent expression of $(1996002)$ in your question, and $(196002)$ in your comment above, I have to wonder whether you might have made a mistake in describing the constraints of the problem that was assigned to you. – user2661923 Apr 17 '22 at 07:50