3. [30] The coefficients of the polynomial $P(x)$ are nonnegative integers, each less than 100 . Given that $P(10)=331633$ and $P(-10)=273373$, compute $P(1)$.
Answer: 100
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
$$
P(1)=a_{0}+a_{1}+a_{2}+\cdots=100 .
$$