3. (CUB 3) ${ }^{\mathrm{IMO} 1}$ Let $n>m \geq 1$ be natural numbers such that the groups of the last three digits in the decimal representation of $1978^{m}, 1978^{n}$ coincide. Find the ordered pair $(m, n)$ of such $m, n$ for which $m+n$ is minimal.
3. What we need are $m, n$ for which $1978^{m}\left(1978^{n-m}-1\right)$ is divisible by $1000=8 \cdot 125$. Since $1978^{n-m}-1$ is odd, it follows that $1978^{m}$ is divisible by 8, so $m \geq 3$. Also, $1978^{n-m}-1$ is divisible by 125, i.e., $1978^{n-m} \equiv 1(\bmod 125)$. Note that $1978 \equiv-2(\bmod 5)$, and consequently also $-2^{n-m} \equiv 1$. Hence $4 \mid n-m=4 k, k \geq 1$. It remains to find the least $k$ such that $1978^{4 k} \equiv 1$ $(\bmod 125)$. Since $1978^{4} \equiv(-22)^{4}=484^{2} \equiv(-16)^{2}=256 \equiv 6$, we reduce it to $6^{k} \equiv 1$. Now $6^{k}=(1+5)^{k} \equiv 1+5 k+25\binom{k}{2}(\bmod 125)$, which reduces to $125 \mid 5 k(5 k-3)$. But $5 k-3$ is not divisible by 5, and so $25 \mid k$. Therefore $100 \mid n-m$, and the desired values are $m=3, n=103$.