Find a three-digit number, any integer power of which ends in three digits that form the original number (in the same order).

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Answer: 376 and 625. Let $N-$ be the desired number. Then $N^{2}-N=N(N-1)$ is divisible by 1000. The numbers $N$ and $N-1$ are coprime, so one of them is divisible by 8, and the other by 125. Let's first assume $N=125 k$. Then $k \leq 8$. Among the numbers $125 k-1, k=1, \ldots, 8$, only the number 624 is divisible by 8. Now let $N-1=125 k$. Then $N=125 k+1$, so $k \leq 7$. Among the numbers $125 k+1, k=1, \ldots, 7$, only the number 376 is divisible by 8. If $N^{2}-N=N(N-1)$ is divisible by 1000, then $N^{\mathrm{k}}-N$ $=N\left(N^{\mathrm{k}-1}-1\right)$ is also divisible by 1000, since $N^{\mathrm{k}-1}-1$ is divisible by $N-1$.