3. Let $m>n \geqslant 1$, find the minimum value of $m+n$ such that: $1000 \mid 1978^{m}-1978^{n}$.
3. Solution: When $n \geqslant 3$, from $1000 \mid 1978^{m}-1978^{n}$, we get
$$
125 \mid 1978^{m-n}-1 \text {, and } 1978=15 \times 125+103 \text {, }
$$

So $125 \mid 103^{m-n}-1$, thus $25 \mid 103^{m-n}-1$.
So $25 \mid 3^{m-n}-1$.
Let $m-n=l$, then $: 25 \mid 3^{l}-1$ (obviously $l$ is even, otherwise it is easy to know $3^{l}-1 \equiv \pm 3-1 \equiv 4$ or $2(\bmod 5)$, which is a contradiction), let $l=2a$,
then $25 \mid 9^{a}-1$, and $9^{a}-1=(10-1)^{a}-1$
$$
\begin{array}{l}
\equiv C_{a}^{2} 10^{2}(-1)^{a-2}+C_{a}^{1} 10(-1)^{a-1}+(-1)^{a}-1(\bmod 25) \\
\equiv 10 a(-1)^{a-1}+(-1)^{a}-1(\bmod 25)
\end{array}
$$

Obviously $a$ is even.
So $9^{a}-1 \equiv 10 a(-1)(\bmod 25)$.
So $5 \mid a$, and $a$ is even, so $10 \mid a$.
Let $a=10b$, then $l=20b$.
So $103^{t}-1=(100+3)^{20b}-1$
$$
\begin{array}{l}
\equiv C_{20b}^{1} 100 \cdot 3^{20b-1}+3^{206}-1(\bmod 125) \\
\equiv 9^{106}-1(\bmod 125) \\
\equiv(10-1)^{10b}-1(\bmod 125) \\
\equiv C_{10b}^{2} \cdot 10^{2}-C_{10b}^{1} 10(\bmod 125) \\
\equiv-100b(\bmod 125),
\end{array}
$$

So $5 \mid b$, so $l=20b \geqslant 100$. So $m+n=l+2n \geqslant 106$.
When $m=103, n=3$ the above can be equal.
If $n \leqslant 2$, it is easy to know $8 \times 1978^{m}-1978^{n} \Rightarrow 1000 \times 1978^{m}-1978^{n}$, so the minimum value of $m+n$ is 106.