\section*{Problem 1 - 181041}

What are the last two digits (in the usual decimal digit notation) of the number \(x\) that satisfies the equation

\[
\log _{13}\left[\log _{12}\left(\log _{11} x\right)\right]=1
\]
}

For the given number \(x\), it holds that \(\log _{12}\left(\log _{11} x\right)=13\), thus \(\log _{11} x=12^{13}\) and therefore

\[
x=11^{12^{13}}
\]

We determine the last two digits of the powers of \(11^{s}(s=1, 2, \ldots)\):

\[
\begin{array}{lccccccccccc}
\text { last digit } & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 & \ldots \\
\text { of } 11^{s} & 11 & 21 & 31 & 41 & 51 & 61 & 71 & 81 & 91 & 01 & \ldots
\end{array}
\]

\begin{tabular}{lllllllllllll} 
of \(11^{s}\) & 11 & 21 & 31 & 41 & 51 & 61 & 71 & 81 & 91 & 01 & \(\ldots\)
\end{tabular}

It follows: If a natural number \(t\) is multiplied by \(11^{10}\), the resulting number has the same last two digits as the number \(t\).

From this, it further follows: If a natural number \(n\) has the last digit \(w\), then \(11^{u}\) has the same last two digits as \(11^{w}\); because with a natural number \(v\) it is \(u=10 v+w\), so \(11^{u}=\left(11^{10}\right) \cdot 11^{w}\) is formed from \(11^{w}\) by multiplying with \(11^{10}\) \(v\) times.

We now determine the last digit of the powers \(12^{y}(y=1,2, \ldots)\):

\[
\begin{array}{llllll}
\text { y } & 1 & 2 & 3 & 4 & \ldots \\
\text { last digit of } 12^{y} & 2 & 4 & 8 & 6 & \ldots
\end{array}
\]

It follows: If a natural number \(z\) that has the last digit 2 is multiplied by \(12^{4}\), the resulting number also has the last digit 2.

From this, it further follows: The number \(u=12^{13}\) has the last digit \(w=2\); because \(12^{13}=\left(12^{4}\right)^{3} .12\) is formed from 12 by multiplying with \(12^{4}\) three times.

Thus, \(x=11^{12^{13}}\) has the same last two digits as \(11^{2}\), i.e., the digits 2, 1 (in this order).