10. (15 points) If all the coefficients of a polynomial are natural numbers, it is called a "natural polynomial". How many different natural polynomials $P(x)$ are there such that $P(2)=20$?
10. For a positive integer $n$, let $A(n)$ denote the number of distinct natural polynomials $P(x)$ that satisfy $P(2) = n$.
It is easy to prove: for any positive integer $m$, we have
$$
\begin{array}{l}
A(2 m+1)=A(2 m) \\
=A(2 m-1)+A(m) .
\end{array}
$$

In fact, for any natural polynomial $P(x)$ that satisfies $P(2)=2 m+1$, since $P(2)$ is odd, the constant term of $P(x)$ must be odd.

Let $Q(x)=P(x)-1$. Then $Q(x)$ is a natural polynomial, and $Q(2)=P(2)-1=2 m$.

Conversely, for any natural polynomial $Q(x)$ that satisfies $Q(2)=2 m$, let $P(x)=Q(x)+1$. Then $P(x)$ is a natural polynomial, and
$$
P(2)=Q(2)+1=2 m+1 \text {. }
$$

Therefore, $A(2 m+1)=A(2 m)$.
For any natural polynomial $P(x)$ that satisfies $P(2)=2 m$, if $P(0) \neq 0$, let $Q(x)=P(x)-1$. Then $Q(x)$ is a natural polynomial, and
$$
Q(2)=P(2)-1=2 m-1 \text {. }
$$

Such polynomials $P(x)$ have $A(2 m-1)$.
If $P(0)=0$, let $P(x)=x Q(x)$, then $Q(x)$ is a natural polynomial, and $2 m=P(2)=2 Q(2)$. Therefore, $Q(2)=m$, and such polynomials $P(x)$ have $A(m)$.
Thus, $A(2 m)=A(2 m-1)+A(m)$.
Therefore, equation (1) holds.
Clearly, $A(2)=2$ (i.e., $P(x)=2, x$), so,
$$
\begin{array}{l}
A(3)=A(2)=2, \\
A(5)=A(4)=2 A(2)=4, \\
A(7)=A(6)=A(3)+A(4)=2+4=6, \\
A(9)=A(8)=A(4)+A(6)=4+6=10, \\
A(11)=A(10)=A(5)+A(8)=4+10=14, \\
A(13)=A(12)=A(6)+A(10)=6+14=20, \\
A(15)=A(14)=A(7)+A(12)=6+20=26, \\
A(17)=A(16)=A(8)+A(14)=10+26=36, \\
A(19)=A(18)=A(9)+A(16)=10+36=46, \\
A(20)=A(10)+A(18)=14+46=60 .
\end{array}
$$