The polynomial question!!

Algebra Level 2

Polynomial $p(n)$ , where $n$ is a natural number, is such that $p(n)=p(n+1)$ and $p(2020)=2020$ .

Find $p\left(2020^{2020} \right)$ .

2 solutions

A Former Brilliant Member
Jul 1, 2020

$p(x)=p(x+1)$

$\Rightarrow$ $p(1)=p(1+1)$

$\Rightarrow$ $p(1)=p(2)$

Similarly, $p(2)=p(3)$

$p(3)=p(4)$

$\Rightarrow$ $p(1)=p(2)=p(3)=........=p(2020)..........=p(2020^{2020})..............................=2020$

$\Rightarrow$ $p(2020^{2020})=2020$

Zakir Husain
Jul 1, 2020

$2020=p(2020)=p(2020+1)=p(2020+1+1)=p(2020+1+1+1)...=p(2020^{2020})$ $\Rightarrow p(2020^{2020})=2020$

You are my first solver.

A Former Brilliant Member - 11 months, 2 weeks ago