G8.4 If $m n p, n m p, m m p$ and $n n p$ are numbers in base 10 composed of the digits $m, n$ and $p$, such that: $m n p-n m p=180$ and $m m p-n n p=d$. Find $d$.
$\begin{array}{l}100 m+10 n+p-(100 n+10 m+p)=180 \\ 100(m-n)-10(m-n)=180 \\ m-n=2 \\ d=m m p-n n p \\ \quad=100 m+10 m+p-(100 n+10 n+p) \\ =110(m-n) \\ =220\end{array}$