$1 \cdot 6$ Let $n$ be an integer. If the tens digit of $n^{2}$ is 7, what is the units digit of $n^{2}$?
[Solution] Let $n=10 x+y$, where $x$ and $y$ are integers, and $0 \leqslant y \leqslant 9$. Thus, we have
$$
\begin{aligned}
n^{2} & =100 x^{2}+20 x y+y^{2} \\
& =20\left(5 x^{2}+x y\right)+y^{2}
\end{aligned}
$$

If the tens digit of $n^{2}$ is the odd number 7, then the tens digit of $y^{2}$ is odd, which leads to
$$
y^{2}=16 \text { or } 36 \text {. }
$$

Therefore, the units digit of $n^{2}$ must be 6.