$$\left[{\frac{{\mathtt{z}}}{\left({\mathtt{2}}\right)}}\right]{\mathtt{\,-\,}}\left[{\frac{{\mathtt{z}}}{\left({\mathtt{8}}\right)}}\right] = {\mathtt{9}}$$
then
$$\left[{\frac{\left({\mathtt{4}}{\mathtt{\,\times\,}}{\mathtt{z}}\right)}{\left({\mathtt{8}}\right)}}\right]{\mathtt{\,-\,}}\left[{\frac{{\mathtt{z}}}{\left({\mathtt{8}}\right)}}\right] = {\mathtt{9}}$$
I'm stuck here:
$$\left[{\frac{\left({\mathtt{3}}{\mathtt{\,\times\,}}{\mathtt{z}}\right)}{\left({\mathtt{8}}\right)}}\right] = {\mathtt{9}}$$
