10.13 A nine-digit number in the form $\overline{a_{1} a_{2} a_{3} b_{1} b_{2} b_{3} a_{1} a_{2} a_{3}}$ is equal to the square of the product of 5 prime numbers, and $\overline{b_{1} b_{2} b_{3}}=2 \overline{a_{1} a_{2} a_{3}} \quad\left(a_{1} \neq 0\right)$. Find this nine-digit number.
(China Shaanxi Province Mathematics Competition, 1979)
[Solution] Since $\overline{b_{1} b_{2} b_{3}}=2 \overline{a_{1} a_{2} a_{3}}$, that is,
$$
b_{1} \cdot 10^{2}+b_{2} \cdot 10+b_{3}=2 a_{1} \cdot 10^{2}+2 a_{2} \cdot 10+2 a_{3} \text {, }
$$

Let $n=\overline{a_{1} a_{2} a_{3} b_{1} b_{2} b_{3} a_{1} a_{2} a_{3}}$. Then
$$
\begin{aligned}
n= & a_{1} \cdot 10^{8}+a_{2} \cdot 10^{7}+a_{3} \cdot 10^{6}+2 a_{1} \cdot 10^{5}+2 a_{2} \cdot 10^{4} \\
& +2 a_{3} \cdot 10^{3}+a_{1} \cdot 10^{2}+a_{2} \cdot 10+a_{3} \\
= & a_{1} \cdot 10^{2}\left(10^{6}+2 \cdot 10^{3}+1\right)+a_{2} \cdot 10\left(10^{6}+2 \cdot 10^{3}+1\right) \\
& +a_{3}\left(10^{6}+2 \cdot 10^{3}+1\right) \\
= & \left(a_{1} \cdot 10^{2}+a_{2} \cdot 10+a_{3}\right)\left(10^{6}+2 \cdot 10^{3}+1\right) \\
= & \overline{a_{1} a_{2} a_{3}} \cdot 1002001 .
\end{aligned}
$$

Since $1002001=(1001)^{2}=(7 \cdot 11 \cdot 13)^{2}$,
we have $n=\overline{a_{1} a_{2} a_{3}} \cdot 7^{2} \cdot 11^{2} \cdot 13^{2}$.
From the given conditions, $\overline{a_{1} a_{2} a_{3}}$ must be the square of the product of two prime numbers.
Because
$$
a_{1} \neq 0, \quad \overline{b_{1} b_{2} b_{3}}<1000,
$$

we have
$$
100 \leqslant \overline{a_{1} a_{2} a_{3}}<500,
$$

which means
$$
10 \leqslant \sqrt{a_{1} a_{2} a_{3}}<23 .
$$

In the set $\{10,11,12,13, \cdots, 21,22,23\}$, only
$$
\begin{array}{l}
10=2 \cdot 5, \quad 14=2 \cdot 7, \quad 15=3 \cdot 5, \\
21=3 \cdot 7, \quad 22=2 \cdot 11 .
\end{array}
$$

can be written as the product of two prime numbers.
Obviously, we cannot take $14,21,22$, otherwise $7^{4}$ or $11^{4}$ would appear in $n$, which is not allowed.
Therefore,
$$
\begin{array}{l}
\sqrt{a_{1} a_{2} a_{3}}=10 \text { or } 15 . \\
\overline{a_{1} a_{2} a_{3}}=100 \text { or } \overline{a_{1} a_{2} a_{3}}=225 \text {. } \\
\overline{b_{1} b_{2} b_{3}}=200 \text { or } \overline{b_{1} b_{2} b_{3}}=450 \text {. } \\
\end{array}
$$

Thus, the required nine-digit number is 100200100 or 225450225.