12 Find a 9-digit number $M$, such that the digits of $M$ are all different and non-zero, and for $m=2,3, \cdots$, 9, the left $m$ digits of $M$ are multiples of $m$.

Find a 9-digit number $M$, such that the digits of $M$ are all different and non-zero, and for $m=2,3, \cdots$, 9, the number formed by the left $m$ digits of $M$ is a multiple of $m$.
12. Let $M=\overline{a_{1} a_{2} \cdots a_{9}}$ be a number that satisfies the given conditions. From the conditions, we know that $a_{5}=5$, and $a_{2}$, $a_{4}$, $a_{6}$, $a_{8}$ are a permutation of $2$, $4$, $6$, $8$. Consequently, $a_{1}$, $a_{3}$, $a_{7}$, $a_{9}$ are a permutation of $1$, $3$, $7$, $9$. Therefore,
$$a_{4}=2 \text { or } 6\left(\text { because } 4 \mid \overline{a_{3} a_{4}}\right),$$

Furthermore,
$$8 \mid \overline{a_{7} a_{8}}$$

Thus,
$$a_{8}=2,6$$

Hence,
$$\left(a_{4}, a_{8}\right)=(2,6),(6,2)$$

By analyzing these two cases further, we can find a number $M=381654729$ that satisfies the conditions.