*1 In the number 3000003, what digit should the hundred's place and the ten-thousand's place digit 0 be replaced with, so that the resulting number is divisible by 13?
Parse $1001=7 \times 11 \times 13$, so $10^{3} \equiv-1(\bmod 13)$. Let the required digits be $x, y$, then we have $n=\overline{30 x 0 y 03}=3 \cdot 10^{6}+x \cdot 10^{4}+y \cdot 10^{2}+3 \equiv 3+3 x+9 y+3=3(2+x+$ $3 y)(\bmod 13)$. Therefore, $13 \mid n$ if and only if $x+3 y+2=13 m$ ( $m$ is a positive integer).

Since $x+3 y+2 \leqslant 9+3 \times 9+2=38$, thus $m=1$ or 2. When $m=1$, from (1) we get $x=8, y=1 ; x=5, y=2 ; x=2, y=3$. When $m=2$, from (1) we get $x=9, y=5$; $x=6, y=6 ; x=3, y=7, x=0, y=8$.

Therefore, we can obtain 7 numbers divisible by 13: $3080103,3050203,3020303,3090503$, $3060603,3030703,3000803$.