1. For natural numbers $m$ and $n$, $n>m>1$. The last three digits of $1978^{m}$ and $1978^{n}$ in decimal notation are the same in order. Find $m$ and $n$ such that $m+n$ is the smallest possible. (Kuba, 6 points)
The condition is satisfied if and only if

$$
1978^{n}-1978^{m}=1978^{m}\left(1978^{n-m}-1\right)
$$

is divisible by 1000, that is, if it is divisible by 8 and 125 separately. Due to the condition $n>m$, the second factor on the right-hand side of $\left({ }^{*}\right)$ is odd, and thus it is divisible by 8 if and only if $m \geq 3$.

For divisibility by 125, first examine the last digit of $1978^{a}$ for $a=1,2, \ldots$. These are $8,4,2,6,8,4,2,6, \ldots$, forming a cycle of four. Thus, $1978^{a}-1$ is divisible by 5 if and only if $a$ is divisible by 4. The last two digits of $1978^{4 b} = \ldots 256^{b}$ form a cycle of five: $56,36,16,96,76,56,36, \ldots$, so $1978^{4 b}-1$ is exactly divisible by 25 if $b$ is a multiple of 5. Finally, examining the last three digits of $1978^{20 c} = \ldots 256^{5 c} = \ldots 776^{c}$, we get a multiple of 125 for the first time when $c=5$.

This means that if $1978^{n-m}-1$ is divisible by 125, then the value of $n-m$ must be at least 100. Comparing this with the previous condition $m \geq 3$, it is immediately clear that the sought values are $m=3$ and $n=103$.

Remark. To solve the problem, we can use the so-called Euler's theorem: if $a$ and $k$ are coprime, and $\varphi(k)$ denotes the number of integers less than $k$ and coprime to $k$, then $a^{\varphi(k)}-1$ is divisible by $k$. (See, for example, Molnár Emil: Matematikai versenyfeladatok, p. 488.) Since $\varphi(125)=100$ and 1978 and 125 are coprime, $1978^{100}-1$ is divisible by 125. However, this does not mean that 100 is the smallest such exponent, although the problem specifically asks for the smallest one.