Problem (Algebra):
Let $T=2030$. Given that $\mathrm{A}, \mathrm{D}, \mathrm{E}, \mathrm{H}, \mathrm{S}$, and $\mathrm{W}$ are distinct digits, and that $\underline{\mathrm{W}} \underline{\mathrm{A}} \underline{\mathrm{D}} \underline{\mathrm{E}}+\underline{\mathrm{A}} \underline{\mathrm{S}} \underline{\mathrm{H}}=T$, what is the largest possible value of $\mathrm{D}+\mathrm{E}$ ?

Solution:
First note that if $T \geq 10000$, then $\mathrm{W}=9$ and $\mathrm{A} \geq 5$. If $T<10000$ and $x$ is the leading digit of $T$, then either $\mathrm{W}=x$ and $\mathrm{A} \leq 4$ or $\mathrm{W}=x-1$ and $\mathrm{A} \geq 5$. With $T=2030$, either $\underline{\mathrm{W}} \underline{\mathrm{A}}=20$


or $\underline{W} \underline{A}=15$. In either case, $\underline{D} \underline{E}+\underline{S} \underline{H}=30$. Considering values of $D+E$, there are three possibilities to consider:

$\mathrm{D}+\mathrm{E}=11: \underline{\mathrm{D}} \underline{\mathrm{E}}=29, \underline{\mathrm{S}} \underline{\mathrm{H}}=01$, which duplicates digits;

$\mathrm{D}+\mathrm{E}=10: \underline{\mathrm{D}} \underline{\underline{E}}=28, \underline{\mathrm{S}} \underline{\underline{H}}=02$ or $\underline{\mathrm{D}} \underline{E}=19, \underline{\mathrm{S}} \underline{\mathrm{H}}=11$, both of which duplicate digits;

$\mathrm{D}+\mathrm{E}=9: \quad \underline{\mathrm{D}} \underline{\mathrm{E}}=27, \underline{\mathrm{S}} \underline{\mathrm{H}}=03$, in which no digits are duplicated if $\underline{\mathrm{W}} \underline{\mathrm{A}}=15$.

Therefore the answer is $\mathbf{9}$.

Final answer: 9