$10 \cdot 771^{2}, 2^{2}, 3^{2}, \cdots, 123456789^{2}$ What is the digit in the units place of the sum?  
(China Junior High School Mathematics League, 1990)
[Solution] Since $123456789=10 \cdot 12345678+9$,
then
$$
\begin{array}{c}
1^{2}+2^{2}+3^{2}+\cdots+123456789^{2} \text {, the unit digit is } \\
=(1+4+9+6+5+6+9+4+1+0) \cdot 12345678+(1+4+
\end{array}
$$
$9+6+5+6+9+4+1)$, the unit digit is $=5 \cdot 8+5$, the unit digit
$$
=5 \text {. }
$$

Thus, the required unit digit is 5.