\section*{Problem 4 - 331034}

\begin{tabular}{ccccc} 
& \(\mathrm{E}\) & \(\mathrm{I}\) & \(\mathrm{N}\) & \(\mathrm{S}\) \\
+ & \(\mathrm{E}\) & \(\mathrm{I}\) & \(\mathrm{N}\) & \(\mathrm{S}\) \\
+ & \(\mathrm{E}\) & \(\mathrm{I}\) & \(\mathrm{N}\) & \(\mathrm{S}\) \\
+ & \(\mathrm{E}\) & \(\mathrm{I}\) & \(\mathrm{N}\) & \(\mathrm{S}\) \\
+ & \(\mathrm{E}\) & \(\mathrm{I}\) & \(\mathrm{N}\) & \(\mathrm{S}\) \\
\hline\(=\) & \(\mathrm{F}\) & \(\mathrm{U}\) & \(\mathrm{N}\) & \(\mathrm{F}\)
\end{tabular}

The cryptogram shown above represents the task of replacing the letters with digits so that a correctly calculated addition problem is formed.

It should also be noted that the rule must be observed that the digit zero must not appear as the leading digit (for E and F).

Identical letters are to be replaced by the same digits, different letters by different digits.

Investigate whether a solution exists! If this is the case, investigate whether different solutions exist and provide each solution!

Hint: Two solutions are considered different if and only if not every letter in one of these solutions is replaced by the same digit as in the other of these solutions.
}

Since the five addends are equal to each other, the sum is a multiple of 5, and thus ends with the digit 0 or 5. Since \(F\) is also the leading digit of the sum, it follows that \(F=5\) and thus \(S\) is odd.

Since there is no carry in the thousands place during the addition, \(5 \cdot E < 10\) must hold, which, due to \(E \neq 0\), immediately implies \(E=1\). Since \(5 \cdot E = F\) holds, there must also be no carry in the hundreds place. Therefore, \(5 \cdot I < 10\) and since \(I \neq E = 1\), \(I = 0\) must hold. Thus, \(\ddot{U} \geq 2\) is the carry obtained from the tens place.

If \(u\) is the carry that results from the addition of the units, then due to \(S \neq 1\), \(S \geq 3\) follows directly, so \(u \geq 1\). On the other hand, \(S \leq 9\), so \(u \leq 4\).

At the tens place, \(5 \cdot N + 4 \geq 5 \cdot N + u = N + 10 \cdot \ddot{U} \geq 20 + N\) or \(4 \cdot N \geq 16\), so \(N \geq 4\).

Case 1: \(N=4\). Then \(5 \cdot N = 20\), so \(u = 4\) (from which \(S = 9\)) and \(\ddot{U} = 2\) follow. We obtain the solution

\[
\begin{aligned}
& \begin{array}{llll}
1 & 0 & 4 & 9
\end{array} \\
& +\quad 100419 \\
& +\quad 10040 \\
& +\begin{array}{llll}
1 & 0 & 4 & 9
\end{array} \\
& \begin{array}{ccccc}
+ & 1 & 0 & 4 & 9 \\
\hline= & 5 & 2 & 4 & 5
\end{array}
\end{aligned}
\]

Case 2: \(N=6\) or \(N=8\). Then \(5 \cdot N + u\) ends with the digit \(u \leq 4 < N\), so no solution exists here.

Case 3: \(N=5\). Then \(N = F\), so no solution exists here.

Case 4: \(N=7\). Then \(5 \cdot N = 35\), so \(u = 2\), from which \(S = 5 = F\) follows, so no solution exists here.

Case 5: \(N=9\). Then \(5 \cdot N = 45\), so \(u = 4\), which leads to the contradiction \(S = 9 = N\).

Thus, there is only the one solution already given for the cryptogram.