I know the range of the $n$ digit numbers in $r$'s complement system is given as
$r^{n-1}-1$ to 0 to $-r^{n-1}$.
So for 3 bit 2's complement, its:
$(2^2-1)$ to $0$ to $-(2^2)$
that is
$3$ to $0$ to $-4$
that is
$(011)_{\text{2's comp}}$ to $(0)_{\text{2's comp}}$ to $(100)_{\text{2's comp}}$
And for 3 digit 8's complement, its
$(8^2-1)$ to $0$ to $-(8^2)$
that is
$63$ to $0$ to $-64$
that is
$(077)_{\text{8's comp}}$ to $(0)_{\text{8's comp}}$ to $(700)_{\text{8's comp}}$
Now I came to know that the $r$'s complement of the minimum number in the range will be the same number.
the $r$'s complement of the smallest / most $-$ve number in the range will be the same number.
For example, taking 2's complement of most $-$ve / smallest number possible with 3 bit 2's complement system, $(100)_{\text{2's comp}}$ is itself.:
1 0 0
0 1 1 (1's complement)
1 0 0 (add 1 to get 2's complement)
But taking 8's complement of most $-$ve / smallest number possible with 3 digit 8's complement system, $(700)_{\text{8's comp}}$ does not seem to be itself:
7 0 0
0 7 7 (7's complement)
1 0 0 (add 1 to get 8's complement)
So 8's complement of 700 does not seem to be 700 itself but is in fact 100. So what I am missing? Or I am doing it all wrong?