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Algebra Level 5

Let $p(x)$ and $q(x)$ be 2557th degree polynomials which satisfy the system of equations:

$\begin{cases} p(n) = q(n)\text{ for }n = 1, 2, 3, \cdots, 2557\\ p(2558) = q(2558) +1. \end{cases}$

Find $p(0) - q(0).$

2 solutions

Ronak Agarwal
Aug 11, 2014

Very simple question. The trick is to assume another polynomial $h(x)=p(x)-q(x)$ . So it is observed that the roots of $h(x)$ are $1,2,3.......2557$

Hence $h(x)=a(x-1)(x-2)(x-3)......(x-2257)$

Also $h(2558)=1 \quad \Rightarrow a=\frac{1}{2557!}$

$h(x)=\frac{(x-1)(x-2)(x-3)......(x-2257)}{2557!}$

$\Rightarrow h(0)=-1$

Nice solution, but a typo. $a=\frac{1}{2557!}$ .

Satvik Golechha - 6 years, 10 months ago

Edited,thanks for reporting

Ronak Agarwal - 6 years, 10 months ago

nice solution

Shubhendra Singh - 6 years, 10 months ago

very nice ....

Vighnesh Raut - 6 years, 9 months ago

Aayush Patni
Jan 3, 2016

I solved using method of differences.

Let f(x)=p(x)-q(x) where f(x) is a polynomial of degree 2557.

Thus the 2557th column in difference table must be constant.

Let f(0)=c

Then using the table

One gets -c=1.

Thus c=-1.