10. (14 points) Given that $p$ is a prime number, $n$ is a positive integer, and non-negative integers $a_{0}, a_{1}, \cdots, a_{n}$ are all less than $p$, and satisfy
$$
\left\{\begin{array}{l}
a_{1}+a_{2}+\cdots+a_{n}=13, \\
a_{0}+a_{1} p+\cdots+a_{n} p^{n}=2015 .
\end{array}\right.
$$

Find the prime number $p$.
10. From the given, we have
$$
\begin{array}{l}
a_{1}(p-1)+a_{2}\left(p^{2}-1\right)+\cdots+a_{n}\left(p^{n}-1\right) \\
=2002 .
\end{array}
$$

Thus, $p-1$ is a positive divisor of $2002=2 \times 7 \times 11 \times 13$.
If $p=2$, then
$$
2015=(11111011111)_{2} \text {, }
$$

which means $\left(a_{n} \cdots a_{1} a_{0}\right)_{2}$ is the binary representation of 2015.
However, $1+1+1+1+1+0+1+1+1+1+1$ $=10 \neq 13$, which is a contradiction.
If $p>2$, then $p$ is odd, and $p-1$ is even.
Therefore, $p-1$
$$
\begin{aligned}
= & 2 \times 7,2 \times 11,2 \times 13,2 \times 7 \times 11,2 \times 7 \times 13, \\
& 2 \times 11 \times 13,2 \times 7 \times 11 \times 13 .
\end{aligned}
$$

Since $p$ is a prime number, $p$ can only be $3, 23, 2003$.
If $p=3$, because $2015=(2202122)_{3}$, so $\left(a_{n} \cdots a_{1} a_{0}\right)_{3}$ is the ternary representation of 2015.
However, $2+2+1+2+0+2+2=11 \neq 13$, which is a contradiction.
If $p=23$, because $2015=3 \times 23^{2}+18 \times 23+14$, so $\left(a_{n} \cdots a_{1} a_{0}\right)_{23}$ is the 23-ary representation of 2015.
However, $14+18+3 \neq 13$, which is a contradiction.
If $p=2003$, because $2015=1 \times 2003+12$, so $\left(a_{n} \cdots a_{1} a_{0}\right)_{2003}$ is the 2003-ary representation of 2015. Notice that, $1+12=13$, which satisfies the given condition. In conclusion, the prime number $p$ is $\mathbf{2 0 0 3}$.