\section*{Problem 3 - 071233}

What are the last two digits of the number \(7^{7^{7^{7}}}-7^{7^{7}}\) ?
}

It is

\[
\begin{array}{lll}
7^{1} \equiv 7 & (\bmod 100) & 7^{2} \equiv 49 \quad(\bmod 100) \\
7^{3} \equiv 43 \quad(\bmod 100) & 7^{4} \equiv 1 \quad(\bmod 100)
\end{array}
\]

thus

\[
7^{4 k} \equiv 1 \quad(\bmod 100) \quad 7^{4 k-1} \equiv 43 \quad(\bmod 100)
\]

where \(k\) is any non-zero natural number. Now,

\[
7 \equiv-1 \quad(\bmod 4), \text { so } \quad 7^{7} \equiv-1 \quad(\bmod 100)
\]

i.e., \(7^{7}=4 m-1\), where \(m\) is a non-zero natural number. It follows that

\[
7^{7^{7}}=7^{4 m-1} \equiv 43 \quad(\bmod 100)
\]

Since

\[
7^{7^{7}} \equiv 7^{4 m-1} \equiv-1 \quad(\bmod 4)
\]

i.e., \(7^{7^{7}}=4 m^{\prime}-1\left(m^{\prime}\right.\) is a non-zero natural \(\left.\mathrm{number}\right)\), it follows further that

\[
7^{7^{7^{7}}}=7^{7^{4 m-1}}=7^{4 m^{\prime}-1} \equiv 43 \quad(\bmod 100)
\]

Therefore, the number to be investigated is

\[
7^{7^{7^{7}}}-7^{7^{7}} \equiv 43-43 \equiv 0 \quad(\bmod 100)
\]

i.e., divisible by 100; each of its last two digits is thus 0.

\section*{Adapted from \([5]\)}