Problem 527

[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.