1. Find all three-digit numbers $n$ whose square ends with the same three-digit number as the square of the number $3 n-2$.
1. If the number $n$ satisfies the condition of the problem, there exists a natural number $k$ such that

$$
(3 n-2)^{2}-n^{2}=1000 k \text {. }
$$

We factor the left side of the given equation using the difference of squares formula and after a simple rearrangement, we get

$$
(2 n-1)(n-1)=250 k=2 \cdot 5^{3} \cdot k \text {. }
$$

The numbers $2 n-1$ and $n-1$ are coprime (since $2 n-1=2(n-1)+1$), and the first one is odd, so the second one must be even. We are looking for odd three-digit numbers $n$ such that either $n-1$ is a multiple of 250, or $2 n-1$ is an odd multiple of the number 125. In the first case, we get $n \in\{251,501,751\}$ and in the second case, we see that $2 n-1 \in$ $\in\{375,625,875,1125,1375,1625,1875\}$, i.e. (since $n$ is odd) $n \in\{313,563,813\}$. In total, the problem has the six solutions mentioned.