long division with no numbers Im stumped by the following brain teaser:
Solve the following long division problem. Each letter represents a unique digit ($0$ - $9$)
$$
\require{enclose}
\begin{array}{r}
                    B  \\[-3pt]
XZD \enclose{longdiv}{BUMG} \\[-3pt]
         \underline{APZK} \\[-3pt]
                    ABU  \\[-3pt]      
\end{array}
$$
im just looking for a first step hint, not the entire solution, or a similar simpler example.

You can separate it into cases on $B$ (it is easy to see that $B\not=0,1,5,9$), and find the values in the following order : $B\rightarrow A\rightarrow X\rightarrow Z,D,K\rightarrow P\rightarrow U,M,G$.
For $B=2$, we have $A=0,1$. But if $A=0$, then 
$$(BU)=(2UMG)-(PZK)\ge 2000-999=1001$$
So, $A=1$. Also, if $X\le 8$, then
$$(2UMG)=2\times (XZD)+(12U)\le 2\times 899+129=1927$$
So, $X=9$.
Here, let $[n]$ be the right-most two digits of $n$. Then, since we have
$$[2\times (ZD)]=[ZK]$$
we have $Z=0$.  But $(2UMG)=2\times (90D)+(12U)\le 2\times 909+129=1947$.
Similarly, you can do for $B=3,4,6,7,8$. In the following, I'll write the outlines.
For $B=3$, we have $A=2,X=9$. Now $Z=4,5$.
Case 1 : If $Z=4$, then $D=7,K=1,P=8,U=0,M=7\quad\Rightarrow\quad D=M$.
Case 2 : If $Z=5$, then $D=0,P=8,K=0\quad\Rightarrow\quad D=K$.
For $B=4$, we have $A=3,X=9$. Now $Z=0,6$.
Case 1 : If $Z=0$, then $D=2,P=6,K=8$. So, $$(4UMG)=4\times 902+(34U)\le 4\times 902+349=3957.$$
Case 2 : If $Z=6$, then $D=5,P=8,K=0,U=2,M=0\quad \Rightarrow\quad K=M$.
For $B=6$, we have $A=5,X=9$. Now $(Z,D,K)=(2k,1,6),(2k-1,9,4)\quad\Rightarrow\quad B=K\quad\text{or}\quad X=D$.
For $B=7$, we have $A=6, X=9$. Now $(Z,D,K)=(3,4,8),(8,3,1)$.
Case 1 : If $(Z,D,K)=(3,4,8)$, then $P=5,U=2,M=1,G=0$. This is sufficient.
Case 2 : If $(Z,D,K)=(8,3,1)$, then $P=8\quad\Rightarrow \quad Z=P$.
For $B=8$, we have $A=7,X=9, Z=1,D=4,K=2,P=3,M=9\quad\Rightarrow\quad M=X$.
Hence, the answer is 
$$A=6,\quad B=7,\quad D=4,\quad G=0,\quad K=8,$$$$\quad M=1,\quad P=5,\quad U=2,\quad X=9,\quad Z=3$$
