10. (20 points) Given
$$
A=\{1,2, \cdots, 2014\}
$$

is a non-empty subset, satisfying that the sum of its elements is a multiple of 5. Find the number of such subsets.
10. Let $A_{j}=\{x \in A \mid x \equiv j(\bmod 5)\}(j=0, 1, \cdots, 4)$. Then
$$
\left|A_{0}\right|=402,\left|A_{j}\right|=403(j=1,2,3,4).
$$

Let $T$ be a non-empty subset that satisfies the condition, and
$$
T \cap A_{j}=T_{j},\left|T_{j}\right|=t_{j}(j=0,1, \cdots, 4).
$$

Then $\sum_{j=0}^{4} j t_{j} \equiv 0(\bmod 5)$.
For each non-negative integer solution $\left(t_{0}, t_{1}, \cdots, t_{4}\right)$ of equation (1) under the conditions $t_{j} \leqslant\left|A_{j}\right|(j=0,1, \cdots, 4)$ and $\sum_{j=0}^{4} t_{j} \neq 0$, the number of corresponding subsets is $\mathrm{C}_{402}^{t_{0}} \prod_{j=1}^{4} \mathrm{C}_{403}^{t_{j 0}}$.
Its generating function is
$$
\begin{array}{l}
f(x)=\left(\sum_{j=0}^{402} \mathrm{C}_{402}^{j} x^{5 j}\right) \prod_{k=1}^{4}\left(\sum_{j=0}^{403} \mathrm{C}_{403}^{j} x^{k j}\right) \\
=\left(1+x^{5}\right)^{402} \prod_{k=1}^{4}\left(1+x^{k}\right)^{403} \\
=\left[\prod_{k=1}^{5}\left(1+x^{k}\right)\right]^{403}\left(1+x^{5}\right)^{-1}.
\end{array}
$$

Therefore, the number of subsets is the sum of the coefficients of all $x^{5 k}$ $\left(k \in \mathbf{N}_{+}\right)$ in $f(x)$, which is $\frac{1}{5} \sum_{j=0}^{4} f\left(\omega^{j}\right)-1$, where $\omega$ is a complex root of $x^{4}+x^{3}+x^{2}+x+1=0$. Thus, $\omega^{5}=1$.
$$
\begin{array}{l}
\text { Also, } f\left(\omega^{0}\right)=f(1)=2^{2014}, \\
f\left(\omega^{j}\right)=\left[\prod_{k=1}^{5}\left(1+\omega^{k j}\right)\right]^{403}\left(1+\omega^{5 j}\right)^{-1} \\
=\frac{1}{2}\left[\prod_{k=0}^{4}\left(1+\omega^{k}\right)\right]^{403}(j=1,2,3,4). \\
\text { Let } g(x)=\prod_{k=0}^{4}\left(x+\omega^{k}\right).
\end{array}
$$

Then $g(x)=0$ has roots $-\omega^{k}(k=0,1, \cdots, 4)$.
Hence, $g(x)=x^{5}+1$.
Therefore, the number of subsets is
$$
\begin{array}{l}
\frac{1}{5} \sum_{j=0}^{4} f\left(\omega^{j}\right)-1=\frac{1}{5}\left[2^{2014}+2(g(1))^{403}\right]-1 \\
=\frac{1}{5}\left(2^{2014}+2^{404}\right)-1.
\end{array}
$$