75. In the six-digit number $M=\overline{a b c 321}$, $a, b, c$ are three different numbers, all greater than 3. If $M$ is a multiple of 7, then the smallest value of $M$ is $\qquad$
Answer: 468321
Solution:
$$
\begin{aligned}
M & =1000 \times \overline{a b c}+321=1001 \times \overline{a b c}-(\overline{a b c}-321) \\
& =7 \times 143 \times \overline{a b c}-(\overline{a b c}-321)
\end{aligned}
$$

Therefore, when $\overline{a b c}-321$ is a multiple of 7, $M$ is a multiple of 7.
Assume $\overline{a b c}-321=7 k$, then
$$
\overline{a b c}=7 k+321 \geq 456,
$$

So
$$
k \geq \frac{135}{7}
$$

When $k=20$,
$\overline{a b c}=140+321=461$ (not meeting the condition),

When $k=21$,
$$
\overline{a b c}=147+321=468,
$$

Therefore, the minimum value of $\mathrm{M}$ is 468321.