This question was asked in a test in my undergrad course
A computer, using 3-digit radix complement arithmetic with an unknown radix r, gives the following results expressed in radix complement form:
(m + n)r = (087)r; (m − n)r = (005)r ; (n − m)r = (184)r
Identify the radix r and the decimal values of m and n.
This is what I did:
Adding last 2 equations,
(m-n+n-m)r = (005)r + (184)r
-> 0 = (005)r + (184)r
-> 0 = 5r0 + 1r2 + 8r1 + 4r0
Because this equation has no integer solutions, therefore no valid r exists.
Am I correct?
Assuming that in $3$-digit $r$'s complement, $-x$ is the complement with respect to $r^3$. Let $x = m-n$
$x = (005)_r$
$-x = (184)_r$
$r^3 = 1\cdot r^2 + 8 \cdot r + 9$
Since $-x$ has a $8$ in it, we have $r\ge 8+1$. But then it will not satisfy our derived equation.
Perhaps there was a typo in the exam?