Find $n>m \geq 1$ that satisfy the condition that the last two digits of the decimal representation of $42^{n}$ are the same as those of $42^{m}$, and such that the sum $m+n$ is minimal.
The statement translates to $42^{n} \equiv 42^{m} \pmod{100}$, or equivalently $100 \mid 42^{m}(42^{n-m}-1)$. Since $100 = 4 \times 25$, the proof is in two steps: finding $n, m$ such that $4 \mid 42^{m}(42^{n-m}-1)$ and the same with 25.

Since $n > m$, $(42^{n-m}-1)$ is odd, so it must be that $4 \mid 42^{m}$, i.e., $m \geq 2$.

Since $42^{m}$ is coprime with 25, it must be that $25 \mid (42^{n-m}-1)$. Fermat's little theorem generalized indicates that this is true if $(n-m) = \phi(25) = 20$, but it is necessary to verify that there is no better solution: all divisors of 20 must be tried.

\[
\begin{aligned}
42^{2} & \equiv (-8)^{2} \equiv 64 \equiv (-11) \pmod{25} \\
42^{4} & \equiv (-11)^{2} \equiv 121 \equiv (-4) \pmod{25} \\
42^{5} & \equiv (-4) \times (-8) \equiv 32 \equiv 7 \pmod{25} \\
42^{10} & \equiv 7^{2} \equiv 49 \equiv (-1) \pmod{25}
\end{aligned}
\]

Therefore, the best solution is $m=2, n=22$ and $m+n=24$.