Why complement of most negative number in octal does not turn out to be itself 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 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?
First of all, if you are considering the $n$th digit of the number as being a sign digit, you can think of it as being either 0 or 1, no matter the value of $r$. In other words, it is always a bit (a  binary digit), and that's how you would represent it in hardware, should that be the case.


  $[1]73_{\text{8's comp}} = -5$
  
  $[1]73_{\text{8's comp}} + [0]06_{\text{8's comp}} = [0]01_{\text{8's comp}}$


The complement of a negative number represented in $r$'s complement will obviously produce the positive version of that same number. The most negative number doesn't have a positive of the same magnitude that fits into the given number of digits (a weakness of this form of representation). 

The number is still there, though, only unsigned:


  $[1]00_{\text{8's comp}} = -64_{10}$
  
  $100_8 = 64_{10}$