1. In the set $M=\{1,2, \cdots, 2018\}$, the sum of elements whose last digit is 8 is $\qquad$
$$
-, 1.204626 .
$$

If all such numbers have their last digit 8 removed, the remaining numbers are $0,1, \cdots, 201$, a total of 202 eights are removed, then the sum sought is
$$
\begin{array}{l}
S=10 \times(0+1+\cdots+201)+8 \times 202 \\
=204626 .
\end{array}
$$