Solve the following system of equations, where $a, b$, and $c$ are bases of number systems and, for example, ${ }^{a)} 101$ represents the number written with digits $1,0,1$ in base $a$.

$$
\begin{aligned}
& { }^{a)} 101+{ }^{b)} 201={ }^{c)} 39+9, \\
& { }^{a)} 203+{ }^{b)} 404={ }^{c)} 77+8 .
\end{aligned}
$$
Since $a, b, c$ represent the base numbers of different number systems, each of them is a positive integer, and based on the digits used, $a \geq 4, b \geq 5, c \geq 10$. Writing out the units of the number systems - for example, ${ }^{a)} 203=2 \cdot a^{2}+0 \cdot a+3-$, we obtain the following system of equations:

$$
\begin{gathered}
\left(a^{2}+1\right)+\left(2 b^{2}+1\right)=(3 c+9)+9 \\
\left(2 a^{2}+3\right)+\left(4 b^{2}+4\right)=(7 c+7)+8
\end{gathered}
$$

After the usual rearrangement,

$$
\begin{aligned}
a^{2}+2 b^{2} & =3 c+16 \\
2 a^{2}+4 b^{2} & =7 c+8
\end{aligned}
$$

we see that this can be considered a system of linear equations in two unknowns, with $a^{2}+2 b^{2}=d$ and $c$ as the unknowns:

$$
d-3 c=16, \quad 2 d-7 c=8
$$

and from this, only $c=24, d=88$ is possible if there is a solution at all, that is,

$$
a^{2}+2 b^{2}=88
$$

From the equation, we can see that $a^{2}$ is even, and therefore divisible by 4:

$$
b^{2}=44-\frac{a^{2}}{2}, \quad \text { where } \frac{a^{2}}{2} \text { is even. }
$$

Since $b^{2}<44$, and $b \geq 5$ and is even, $b$ can only be 6, and then $a=4$.

Thus, the bases of the number systems are $a=4, b=6$, and $c=24$. Substituting back, we can verify the correctness of the solution.