TITLE: Finding $f(100)$ given that $f$ is a polynomial of degree $98$ and $f (k)=\frac 1 k$ for $k=1,2,3,\dots,98,99$
QUESTION [5 upvotes]: A polynomial of degree $98$ such $f (k)=\frac 1 k$ for $k=1,2,3,\dots,98,99$  exists.
How can I find $f(100)$?
What are the possible methods?

REPLY [6 votes]: This is based on Akiva Weinberger columbus's hint. 
Consider the polynomial $$kf(k)-1$$
which has degree 99 is zero for all $k=1,2,3, \cdots, 99$. 
So you know the polynomial $kf(k)-1$ has the form $$C(k-1)(k-2)(k-3)\cdots(k-99)$$
and you can find $C$ when you put $k=0$. 
In general, its a good idea to find a polynomial that has as many zeros as its degree is.