, 2019

Solution:
Let
$$P(x) = a_{0} + a_{1} x + a_{2} x^{2} + \ldots$$
Then
$$\frac{1}{2}(P(10) + P(-10)) = a_{0} + 100 a_{2} + \ldots$$
and
$$\frac{1}{2}(P(10) - P(-10)) = 10 a_{1} + 1000 a_{3} + \ldots$$
Since all the coefficients are nonnegative integers, these expressions give us each of the coefficients by just taking two digits in succession. Thus we have $a_{0} = 3$, $a_{1} = 13$, $a_{2} = 25$, $a_{3} = 29$, $a_{4} = 30$ and $a_{n} = 0$ for $n > 4$. Thus
$$P(1) = a_{0} + a_{1} + a_{2} + \cdots = 100$$