jmc

Looking at the $d$'s place, we see that $A_d+A_d=16_d=d+6$ or $A_d+A_d+1=16_d=d+6$ (if there is carry-over). Re-arranging and solving for $A_d$, we find that $A_d=\frac{d+6}{2}$ or $A_d=\frac{d+5}{2}$. In either case, since $d>6$, it follows that $A_d>2$. Thus, when we add the units digits $B_d+A_d$, there must be carry-over, so $A_d=\frac{d+5}{2}$. It follows that $$B_d+A_d=d+2⟹B_d=d+2-\frac{d+5}{2}=\frac{d}{2}-\frac{1}{2}.$$Thus, $A_d-B_d=\frac{d+5}{2}-\frac{d-1}{2}={\boxed{3}}_d$.