Say you're performing Lagrange Interpolation on a function $P(x)$ and you've found that
$$ P(x) = \sum_{i = 1}^{11} \Delta_i(x) $$
given the eleven points $(1, P(1)), (2, P(2)), ..., (11, P(11))$. Remember that
$$ \Delta_i(x) = \frac{\prod_{i \neq j} (x - x_j)}{\prod_{i \neq j} (x_i - x_j)} $$
What is $P(20)$?
Assuming there is no typo in the question, $P(1)=P(2)=\dots=P(11)=1$ and the interpolation polynomial $P$ has degree 10. Thus, the 10th degree polynomial $R(x)=P(x)-1$ has at least 11 zeros. Since a nonzero polynomial of degree $n$ has at most $n$ zeros, we must have $R(x)=0$. Therefore, $P(x)=1$ for all $x$.