The question in the school magazine : $DBSPD-WILL=WIN$ When I was reading my school magazine, I found some hard problems in the Math Corner page. I can't solve two of the questions. Here is one of them

In the equation below, each letter represents distinct digits from 0 to 9.$$DBSPD-WILL=WIN$$Find the value for each letter.

Can someone help me solve this?
 A: Writing out the subtraction
$$\begin{array}
&D&B&S&P&D\\
-&W&I&L&L\\
\hline
&&W&I&N\end{array}$$
we can trivially observe that $D=1$ and $W=9$, which in turn forces $B=0$:
$$\begin{array}
&1&0&S&P&1\\
-&9&I&L&L\\
\hline
&&9&I&N\end{array}$$
If the rightmost column did not carry from the one to its left, we would have $1-L=N$, which is impossible since $0,1$ are already used. Thus, we have a carry and the relation $L+N=11$.
Similarly, since the left two columns read $10-9=0$, it must have provided a carry to the third column and $10+S-I=9$, or $I=S+1$. (The third column itself cannot provide a carry to the fourth, because that would lead to $S=I$.)
Finally, the fourth column, after accounting for the carry given to the fifth, reads $P-1-L=I$ or $I+L+1=P$.
We now list the possible values for $I$ and $L$, based on the restriction that each variable must now lie between $2$ and $8$ inclusive:
$$I\in\{3,4,5,6,7\}$$
$$L\in\{3,4,5,6,7,8\}$$
The minimum value of $I+L$ is $7$. This is also the maximum, since any higher value would cause $P$ to be $9$ or two digits. Hence $\{I,L\}=\{3,4\}$ and $P=8$, the latter value ruling out $L=3$ since that implies $N=8$. So $L=4$, $N=7$, $I=3$ and $S=2$.
The solved equation is
$$\begin{array}
&1&0&2&8&1\\
-&9&3&4&4\\
\hline
&&9&3&7\end{array}$$
$$10281-9344=937$$
A: You can consider addition instead:
$$\begin{array}{ccc}
&W&I&L&L\\
+&&W&I&N\\
\hline
D&B&S&P&D\end{array}$$
Since $L+N\le 18 \Rightarrow L+I\le 19 \Rightarrow I+W\le 19 \Rightarrow W+1=9+1=10 \Rightarrow \color{red}{W=9}, \color{red}{D=1}, \color{red}{B=0}$. 
Hence:
$$\begin{array}{ccc}
&9&I&L&L\\
+&&9&I&N\\
\hline
1&0&S&P&1\end{array}$$
Since $L,N\not\in \{0,1\}$, $L+N=11$. It means $L+I+1=P \ \text{or} \ P+10$, but it can not be $P+10$, because $I+9+1=S \iff I+10=S \Rightarrow I=S$. So, $L+I+1=P$.
Note $I\ne 0,1,2$, so $I\ge 3$. Similarly, $L\ge 3$. So, $\color{red}{P=8}$. So, two options: $(I,L)=(3,4),(4,3)$, but only the first suits (why?):
$$\begin{array}{ccc}
&9&3&4&4\\
+&&9&3&7\\
\hline
1&0&2&8&1\end{array}$$
