\section*{Exercise 2 - 310912}

Werner is working on creating cryptograms in the form of addition problems. In such a cryptogram, identical letters should be replaced by identical digits and different letters by different digits, so that a correctly calculated addition is then present.

Werner considers the following three cryptograms:

\begin{tabular}{cccccc} 
& \(J\) & \(A\) & \(C\) & \(K\) & \(E\) \\
+ & & \(H\) & \(O\) & \(S\) & \(E\) \\
\hline\(=\) & \(A\) & \(N\) & \(Z\) & \(U\) & \(G\)
\end{tabular}

![](https://cdn.mathpix.com/cropped/2024_06_06_b85a894aa8dbf722a7b3g-1115.jpg?height=134&width=374&top_left_y=2417&top_left_x=864)

\begin{tabular}{ccccc} 
& & \(M\) & \(I\) & \(R\) \\
+ & \(E\) & \(M\) & \(I\) & \(R\) \\
\hline\(=\) & \(R\) & \(E\) & \(I\) & \(M\)
\end{tabular}

Determine for each of these three cryptograms whether it has a solution, and if so, find all solutions of the respective cryptogram!
}
a) This cryptogram has no solution, as it contains more than 10 different letters.

b) Assuming a substitution of letters by digits is a solution to the cryptogram, then for it:

It is \(N \neq 0\), since \(N=0\) would lead to the contradiction \(U=R\) in the units place.

Thus, the addition in the tens place of \(N, A\) and any potential carry can only result in a number with the last digit \(A\) if there is a carry from the units place and \(N=9\).

This leads to a carry from the tens place to the hundreds place, and the addition in the hundreds place can only result in a number with the last digit \(A\) if \(R=9\) as well. This assumption of a substitution being a solution leads to a contradiction; therefore, the cryptogram has no solution.

c) I) Assuming a substitution is a solution to the cryptogram, then for it:

At the thousands place, there is a carry from the hundreds place because \(E \neq R\); it holds that \(M+M \geq 9 \quad\) (1) and \(E+1=R \quad\) (2).

From (1) it follows that \(M>4\); furthermore, it is evident from the units place that \(M\) is even, so \(M=6\) or \(M=8\) must be true. If one performs the addition at the hundreds place for these values (either without or with a carry from the tens place) and then applies (2), one obtains: There are at most the following possibilities:

\begin{tabular}{ccc}
\(M\) & \(E\) & \(R\) \\
\hline 6 & 2 & 3 \\
6 & 3 & 4 \\
8 & 6 & 7 \\
8 & 7 & 8
\end{tabular}

However, the second, third, and fourth possibilities contradict the information at the units place. The fourth also contains the contradiction \(M=R\). Therefore, only the first possibility can hold. In this case, there is no carry from the units place to the tens place; thus, the number \(I+I\) has the last digit \(I\), which is only true for \(I=0\).

Therefore, only the substitution \(M=6, E=2, R=3, I=0\) can solve the cryptogram.

II) It satisfies the equality and inequality conditions, and the addition

![](https://cdn.mathpix.com/cropped/2024_06_06_b85a894aa8dbf722a7b3g-1116.jpg?height=136&width=329&top_left_y=1688&top_left_x=932)

is correctly calculated.

Thus, it is shown that the cryptogram has exactly this solution.