Let $23 a$ be a nine-digit number of the form $\overline{a_{1} a_{2} a_{3} b_{1} b_{2} b_{3} a_{1} a_{2} a_{3}}$, which is equal to the square of the product of five different prime numbers, and $\overline{b_{1} b_{2} b_{3}}=2 \cdot \overline{a_{1} a_{2} a_{3}}$. Find this nine-digit number.
Let $N=\overline{a_{1} a_{2} a_{3} b_{1} b_{2} b_{3} a_{1} a_{2} a_{3}}$. Since
$$
\overline{b_{1} b_{2} b_{3}}=2 \cdot \overline{a_{1} a_{2} a_{3}}<1000,
$$
(both different from $7,11,13$) the square of the product. From (1), we know $\overline{a_{1} a_{2} a_{3}}<500$, so $10 \leqslant \sqrt{a_{1} a_{2} a_{3}}<23$. Between 10 and 22, only 10 and 15 meet the requirements. Therefore, $\overline{a_{1} a_{2} a_{3}}=100$ or 225. Thus, the nine-digit number we are looking for is 100200100 or 225450225.