\section*{Problem 1 - 161011}

In the cryptogram, digits \((0,1, \ldots, 9)\) are to be inserted in place of the letters so that a correct addition problem is formed. The same letters should represent the same digits, and different letters should represent different digits.

\begin{tabular}{rrrrr}
\(\mathrm{K}\) & \(A\) & \(T\) & \(E\) & \(R\) \\
\(+\quad \mathrm{K}\) & \(\mathrm{A}\) & \(\mathrm{T}\) & \(\mathrm{Z}\) & \(\mathrm{E}\) \\
\hline \(\mathrm{T}\) & \(\mathrm{I}\) & \(\mathrm{E}\) & \(\mathrm{R}\) & \(\mathrm{E}\)
\end{tabular}

Provide all solutions to this problem!
}

Imagine the columns of the problem numbered from left to right. Then:

(1) From \(r+E=E\) it follows that \(R=0\) (Column 5).

(2) Since \(R \neq 0\) and therefore \(E \neq 0\), it follows that \(E+z=10\) (Column 4).

(3) Since \(E \neq z\) and (2), it follows that \(E \neq 5, Z \neq 5\).

(4) From (2), it follows that \(2 T+1=E\) or \(2 T+1=10+E\) (Column 3).

(5) From (3) and (4), it follows that \(T \neq 2\) and \(T \neq 7\).

(6) From (1), it follows that \(K>0\) and thus \(T \neq 7\) (Column 1).

(7) From (4) and \(T \neq E\), it follows that \(T \neq 9\).

Thus, \(T\) can only be one of the numbers \(3, 4, 5, 6\) or 8.

(I) Let \(T=3\). Then \(E=7\) and \(Z=3\), which contradicts \(Z \neq T\). Therefore, \(T \neq 3\).

(II) Let \(T=4\). Then \(E=9\) and \(Z=1\). Since \(T\) is even, there can be no carry from Column 2 to Column 1. Thus, \(K=2\) and \(A<5\), which, considering the already used numbers, leads to \(A=3\) and \(I=6\). This results in the following solution:

\[
\begin{array}{r}
2344190 \\
+\quad 234 \\
\hline 4690
\end{array}
\]

(III) Let \(T=5\). Then \(E=1\), \(Z=9\), and \(K=2\), since a carry from Column 2 to Column 1 must occur for odd \(T\). Since \(T=5\), \(2 A+1=10+I\), so considering the already used number, \(I=7\) and thus \(A=8\). This results in another solution:

![](https://cdn.mathpix.com/cropped/2024_06_06_b85a894aa8dbf722a7b3g-1373.jpg?height=150&width=385&top_left_y=1790&top_left_x=884)

(IV) Let \(T=6\). Then \(E=K=3\), which contradicts \(E \neq K\). Therefore, \(T \neq 6\).

(V) Let \(T=8\). Then \(E=7\), \(Z=3\), and \(K=4\), since in this case there can be no carry from Column 2 to Column 1. From \(I=2 A+1\) and considering the already used numbers, it follows that \(I=5\) and \(A=2\). This results in the third solution to the problem:

\[
\begin{array}{r}
428870 \\
+\quad 42883 \\
\hline 857007
\end{array}
\]

The given solutions are also all possible solutions.