7. Determine the number of 8-digit positive integers such that after deleting any one digit, the remaining 7-digit number is divisible by 7 .
7. Answer: 64
Let $N=\overline{a b c d e f g h}$ be such a number. By deleting $a$ and $b$, we get $\overline{b c d e f g h}$ and $\overline{\operatorname{acdef} g h}$ respectively. Both of them are divisible by 7 , hence their difference $1000000(b-a)$ is also divisible by 7 , therefore $b-a$ is divisible by 7. By the similar argument, $c-b, d-c, e-d, f-$ $e, g-f, h-g$ are divisible by 7. In other word, all the digits of $N$ are congruent modulo 7 . If $N$ contains digits that are greater than 7 , then one can subtract 7 from each digit to get a new number $N$. Then $N$ satisfies the requirements in the question if and only if $N$ does. Since all the digits are congruent modulo 7 , it remains to consider numbers of the form pppppppp, where $p=0,1,2,3,4,5,6$. By deleting a digit in this number, we get the number $\overline{p p p p p p p}=1111111 p$, which is divisible by 7 if and only if $p$ is the digit 0 or 7 . However, the first two digits of $N$ must be 7 since the number $N$ has 8 digits and that any number we get by deleting a digit in $N$ has 7 digits. On the other hand, the remaining 6 digits can be independently 0 or 7 . Consequently, there are $2^{6}=64$ choices of such numbers.