Inspired by Chan Lye Lee

We know that $\sum_{k=0}^{101}(-1)^k{101 \choose k} f(k)=0$ when $\deg(f)<101$ and $\sum_{k=0}^{101}(-1)^k{101 \choose k}2^k$ $=-(2-1)^{101}$ $=-1$ . Since all the summands are the same except for $k=2$ , it follows that ${101 \choose 2}f(2)={101 \choose 2}4+1$ and $f(2)=4+\frac{1}{5050}$ . The answer is $\boxed{5050}$ .

The polynomial $u(x) \; = \; \sum_{m=0}^{98} \binom{x}{m}$ has degree $98$ and is such that $u(m) = 2^m$ for $0 \le m \le 98$ . Thus $g(x) = 8u(x-3)$ has degree $98$ and is such that $g(m) = 2^m$ for $3 \le m \le 101$ . Thus the polynomial we want is of the form $f(x) \; = \; (Ax + B)\prod_{j=3}^{101}(j-x) + g(x) \;,$ where $A$ and $B$ are constants chosen so that $1 \; = \; f(0) \; = \; \tfrac{101!}{2}B + g(0) \qquad \qquad 2 \; = \; f(1) \; = \; (A + B)100! + g(1) \;.$ Thus $f(2) \; = \; (2A+B)99! + g(2) \; = \; \frac{402 - 202g(1) + 2g(0)}{10100} + g(2) \;,$ With $g(0) = 20000$ , $g(1) = 400$ and $g(2) = 8$ , we obtain $f(2) = \tfrac{20201}{5050}$ , and hence the answer is $\tfrac{1}{f(2) - 4} \,=\, \boxed{5050}$ .