6. Find the ordered quadruple of digits $(A, B, C, D)$, with $A>B>C>D$, such that
$$
\begin{aligned}
& A B C D \\
-\quad & D C B A \\
= & B D A C .
\end{aligned}
$$
Solution: $(7,6,4,1)$
Since $D<A$, when $A$ is subtracted from $D$ in the ones' column, there will be a borrow from $C$ in the tens' column. Thus, $D+10-A=C$. Next, consider the subtraction in the tens' column, $(C-1)-B$. Since $C<B$, there will be a borrow from the hundreds' column, so $(C-1+10)-B=A$. In the hundreds' column, $B-1 \geq C$, so we do not need to borrow from $A$ in the thousands' column. Thus, $(B-1)-C=D$ and $A-D=B$. We thus have a system of four equations in four variables $A, B, C, D$, and solving by standard methods (e.g. substitution) produces $(A, B, C, D)=(7,6,4,1)$.