Question 220: Among the 10-digit decimal numbers formed by the digits $3, 4, 5, 6$, how many are divisible by 66667?
Question 220, Solution: Let the ten-digit number that satisfies the conditions be $\overline{\mathrm{abcdepqrst}}$. Then $3 \leq$
$a, b, c, d, e, p, q, r, s, t \leq 6$.
Let $\overline{a b c d e p q r s t}=66667 \mathrm{k}$, then $\overline{a b c d e} \cdot 10^{5}+\overline{\text { pqrst }}=66666 \mathrm{k}+\mathrm{k}$, that is,
$\overline{\mathrm{pqrst}}+\overline{\mathrm{abcde}}-\mathrm{k}=33333(2 \mathrm{k}-3 \overline{\mathrm{abcde}})$
Also, because $3333333333 \leq \overline{\text { abcdepqrst }} \leq 6666666666$, and $\left|\frac{3333333333}{66667}\right|=50000$,
$\left[\frac{6666666666}{6666}\right]=99999$, so $50000 \leq \mathrm{k} \leq 99999$. Also, $33333 \leq$ pqrst $\leq 66666,33333 \leq$ abcde $\leq 66666$. Therefore,
$\overline{\text { pqrst }}+\overline{\text { abcde }}-\mathrm{k} \geq 33333+33333-99999=-33333$
$\overline{\text { pqrst }}+\overline{\text { abcde }}-\mathrm{k} \leq 66666+66666-50000<3 \times 33333$

Thus, according to $(*)$, it can only be that $2 \mathrm{k}-3 \overline{\mathrm{abcde}}=-1, 0, 1$ or 2.
(1) If $2 \mathrm{k}-3 \overline{\mathrm{abcde}}=-1$, according to ( $*$ ) we know $\mathrm{k}=\overline{\mathrm{pqrst}}+\overline{\mathrm{abcde}}+33333$. Therefore, 3 $\overline{\text { abcde }}-1=2 \times \overline{\text { pqrst }}+2 \times \overline{\text { abcde }}+2 \times 33333$
$\Rightarrow \overline{\text { abcde }}=2 \times \overline{\text { pqrst }}+66667$
This is impossible.
(2) If $2 \mathrm{k}-3 \overline{\mathrm{abcde}}=0$, according to ( $*$ ) we know $\mathrm{k}=\overline{\mathrm{pqrst}}+\overline{\mathrm{abcde}}$, thus
$3 \times \overline{\text { abcde }}=2 \times \overline{\text { pqrst }}+2 \times \overline{\text { abcde }}$
$\Rightarrow \overline{\mathrm{abcde}}=2 \times \overline{\mathrm{pqqst}}$

At this point, only $\overline{\mathrm{pqrst}}=33333, \overline{\mathrm{abcde}}=66666$, i.e., $\overline{\mathrm{abcdepqrst}}=6666633333$.
(3) If $2 \mathrm{k}-3 \overline{\mathrm{abcde}}=1$, according to $(*)$ we know $\mathrm{k}=\overline{\mathrm{pqrst}}+\overline{\mathrm{abcde}}-33333$, thus
$3 \times \overline{\text { abcde }}+1=2$ (pqrst $+\overline{\text { abcde }}-33333$ )
$\Rightarrow \overline{\text { abcde }}+1=2 \times($ pqrst -33333 )
Let $\overline{\mathrm{pqrst}}-33333=\overline{\mathrm{xyzmn}}$, then $0 \leq x, y, z, m, n \leq 3$, and $\overline{\mathrm{abcde}}+1=2 \times \overline{\mathrm{xyzmn}}$.

Clearly, $\overline{a b c d e}+1$ and $2 \cdot \overline{x y z m n}$ do not involve any carry in the calculation, so each digit of $2 \times \overline{x y z m n}$ is between
3 and 6, so $x, y, z, m, n$ each digit is 2 or 3. On the other hand, when $x, y, z, m, n$ each digit is 2 or 3, they all satisfy the problem's conditions. This results in $2^{5}=32$ ten-digit numbers that satisfy the problem's conditions.
(4) If $2 \mathrm{k}-3 \overline{\mathrm{abcde}}=2$, according to (*) we know $\mathrm{k}=\overline{\mathrm{pqrst}}+\overline{\mathrm{abcde}}-66666$. Thus
$3 \times \overline{\mathrm{abcde}}+2=2(\overline{\text { pqrst }}+\overline{\mathrm{abcde}}-66666)$
$\Rightarrow \overline{\text { abcde }}+2 \times 66666+1=2 \times \overline{\text { pqrst }}$

This is impossible.
In summary, there are $32+1=33$ ten-digit positive integers that satisfy the problem's conditions.