$10 \cdot 82$ number $1978^{n}$ and $1978^{m}$ have the same last three digits. Try to find the positive integers $m$ and $n$, such that $m+n$ takes the minimum value (here $n>m \geqslant 1$).
（20th International Mathematical Olympiad, 1978）
[Solution] Since the last three digits of $1978^{n}$ and $1978^{m}(n>m \geqslant 1)$ are the same, we have
$$
1978^{n}-1978^{m}=1978^{m}\left(1978^{n-m}-1\right)
$$

is a multiple of $1000=2^{3} \cdot 5^{3}$.
Since $1978^{n-m}-1$ is an odd number, it has no factor of 2, so from (1), $1978^{m}$ must be divisible by $2^{3}$. Since $1978^{m}=2^{m} \cdot 989^{m}$, and 989 is an odd number, we have $m \geqslant 3$.
Since $1978^{m}$ has no factor of 5, from (1), $1978^{n-m}-1$ must be divisible by $5^{3}$.
Notice that $1978^{t}$, when $t=1,2,3, \cdots$, its unit digit cycles through $8,4,2,6,8,4,2$, $6, \cdots$, so only when $n-m=4 k$ does $1978^{n-m}-1$ possibly be divisible by $5^{3}$.
Since
$$
1978^{4 k}=(2000-22)^{4 k},
$$

to make $1978^{4 k}-1$ divisible by $5^{3}$, it is only necessary for $22^{4 k}-1$ to be divisible by $5^{3}$. And
$$
\begin{array}{l}
22^{4 k}=484^{2 k}=(500-16)^{2 k} . \\
16^{2 k}=256^{k}=(250+6)^{k} .
\end{array}
$$

Thus, it is only necessary for $6^{k}-1$ to be divisible by $5^{3}$.
$$
\begin{aligned}
6^{k}-1 & =(6-1)\left(6^{k-1}+6^{k-2}+\cdots+6+1\right) \\
& =5\left(6^{k-1}+6^{k-2}+\cdots+6+1\right) .
\end{aligned}
$$

Therefore, from (2), $6^{k-1}+\cdots+6+1$ must be divisible by $5^{2}=25$.
Since $6^{t}$ leaves a remainder of 1 when divided by 5, and $6^{k-1}+\cdots+6+1$ has $k$ terms, $k$ should be a multiple of 5, let $k=5 p$.
$$
\begin{aligned}
6^{k}-1 & =6^{5 p}-1 \\
& =7776^{p}-1 \\
& =(7776-1)\left(7776^{p-1}+7776^{p-2}+\cdots+7776+1\right) .
\end{aligned}
$$

Since $7776-1=7775=311 \cdot 25$ is a multiple of 25, and 7776 leaves a remainder of 1 when divided by 5, to make $7776^{p-1}+\cdots+7776+1$ divisible by 5, the minimum value of $p$ is 5, the minimum value of $4 k$ is 100, that is, the minimum value of $n-m$ is 100.
Since
$$
n+m=(n-m)+2 m,
$$

so when $m$ takes the minimum value of 3, and $n-m$ takes the minimum value of 100, at this time $n=103$, $n+m$ has the minimum value of 106.