We are given a polynomial of degree $9$ and for all $k \in \{1, 2, ..., 10\}$ we have $f(k)=k$. Find value of $f(100)$.
My approaches were to solve the system of equations to have full formula for polynomial or use Lagrange Interpolation Polynomial. However, these are tedious ways of dealing with this problem. Am I not forgetting any theorem about polynomials that would be helpful?
On
I assume you meant to say degree 9 (since otherwise there are infinitely many possible values of $f(100)$). If so, use the fact that the Lagrange interpolation polynomial is unique: what's the simplest function you know where $f(x)=x$ for a large number of $x$? Does it happen to be a polynomial of degree (at most) 10?
Define $g (x)=f (x)-x$. Then $g $ is still a degree 10 polynomial, but now $g (k)=f (k)-k=0$. Therefore $g (x)=(x-1)(x-2)\ldots (x-9)(x-10) $, and then $g (100)=99(98)(97)\ldots (90)$, so that $f (100)=99 (98)\ldots (90)+100$.