\section*{Problem 4 - 151014}

In

\[
\begin{array}{ccccc} 
& \mathrm{H} & \mathrm{A} & \mathrm{U} & \mathrm{S} \\
+ & \mathrm{H} & \mathrm{A} & \mathrm{U} & \mathrm{S} \\
+ & \mathrm{H} & \mathrm{A} & \mathrm{U} & \mathrm{S} \\
\hline \mathrm{S} & \mathrm{T} & \mathrm{A} & \mathrm{D} & \mathrm{T}
\end{array}
\]

the letters should be replaced by digits so that a correctly solved addition problem is formed. The same letters should be replaced by the same digits, and different letters should be replaced by different digits.

Provide all solutions!
}

The columns are labeled from the units place as 1 to 5.

I. Suppose that a substitution leads to a solution. Then for this substitution, \(S\) can only be one of the digits 1 or 2, since the sum of three four-digit numbers is less than 30000, and the five-digit natural number "STADT" cannot start with 0.

Case 1: Let \(S=1\).

From column 1, it immediately follows that \(T=3\).

Since the sum starts with the digits 1 and 3, and among the numbers that are equal to 13 or 1 or 2 less than 13, only \(12=3 \cdot 4\) is divisible by 3, only \(H=4\) is possible, and a carry of 1 must be generated from column 3.

Thus, \(A\) must represent one of the digits for which the triple or the triple increased by 1 or 2 is one of the numbers \(10, \ldots, 19\). Of all such digits 3, 4, 5, or 6, 3 and 4 are excluded because they are already assigned. Since a carry of at most 2 can be generated from column 2 and \(3 \cdot 6=18\), adding 0, 1, or 2 does not result in a number ending in 6, so \(A=6\) is excluded.

Therefore, \(A\) can only be 5, and for the sum in column 3 to end in 5, column 2 must not generate a carry, so \(U\) can only be one of the digits \(0, 1, 2\), or 3. Of these, 1 and 3 are already assigned.

For \(U=0\), \(D=0\) would result, leading to a contradiction. Thus, \(U\) can only be 2, from which it follows that \(D=6\).

Case 2: Let \(S=2\).

From column 1, it immediately follows that \(T=6\).

Since the sum starts with the digits 2 and 6, and among the numbers that are equal to 26 or 1 or 2 less than 26, only \(24=3 \cdot 8\) is divisible by 3, only \(H=8\) is possible, and a carry of 2 must be generated from column 3. Thus, \(A\) must represent one of the digits for which the triple or the triple increased by 1 or 2 is one of the numbers \(20, \ldots, 29\).

Of all such digits, 6, 7, 8, or 9, 6 and 8 are excluded because they are already assigned. Since a carry of at most 2 can be generated from column 2 and \(3 \cdot 7=21\), adding 0, 1, or 2 does not result in a number ending in 9, so \(A=7\) is excluded.

Therefore, \(A\) can only be 9, and for the sum in column 3 to end in 9, a carry of 2 must be generated from column 2.

Thus, \(U\) can only be one of the digits 7, 8, or 9. Of these, 7 is the only one not yet assigned, from which it follows that \(D=1\).

Therefore, the substitutions \(A=5, D=6, H=4, S=1, T=3, U=2\), and \(A=9, D=1, H=8, S=2, T=6, U=7\) can lead to solutions.

II. Indeed, with these substitutions, different letters are always replaced by different digits, and since the resulting addition problems

\begin{tabular}{|c|c|c|c|c|c|c|c|c|}
\hline & 4 & 5 & 2 & 1 & & 8 & 9 & 7 \\
\hline+ & 4 & 5 & 2 & 1 & + & 8 & 9 & 7 \\
\hline & 4 & 5 & 2 & 1 & + & 8 & 9 & 7 \\
\hline\(\overline{1}\) & 3 & 5 & 6 & 3 & \(\overline{2}\) & 6 & 9 & 1 \\
\hline
\end{tabular}

are correctly solved, these substitutions satisfy all the given requirements.

Solutions of the 1st Round 1975 taken from [5]

\subsection*{7.17.2 2nd Round 1975, Class 10}