Find all 3-digit integers $n$ such that the base 10 representation of $x^{2}$ ends with the base 10 representation of $x$.

For example, for 1 digit, we have $5^{2}=25,6^{2}=36$, and for 2 digits, we have $25^{2}=625$ and $76^{2}=5776$.
We are looking for all $x$ such that:

$$
1000 \mid x^{2}-x=x(x-1)
$$

As usual, we break down the problem into the different prime factors. We are therefore looking for all $x$ such that:

$$
\begin{array}{r}
125 \mid x(x-1) \\
8 \mid x(x-1)
\end{array}
$$

Since $x$ is coprime with $x-1$, we have:

$$
x \equiv 0 \text { or } 1[125] \quad x \equiv 0 \text { or } 1[8]
$$

By applying the Chinese Remainder Theorem (CRT), we find the $x$ that work modulo 1000. For this, we need to calculate the inverse of 125 modulo 8, and the inverse of 8 modulo 125, which are 5 and 47, respectively. Thus, the solutions are:

$$
125 \cdot 5=625 \text { and } 8 \cdot 47=376
$$