3-2. Find a three-digit number, every integer power of which ends in the same three digits as the original number (in the same order).
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 $k \leqslant 8$. 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 $k \leqslant 7$. 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 $N^{k}-N=N\left(N^{k-1}-1\right)$ is also divisible by 1000, since $N^{k-1}-1$ is divisible by $N-1$.