1、In the set $M=\{1,2, \cdots, 2018\}$, the sum of elements whose last digit is 8 is $\qquad$Note: 204626.
Solution: If all such numbers have their last digit 8 removed, the remaining numbers are $0,1,2, \cdots, 201$, a total of 202 8s are removed, so the sum we are looking for is $S=10 \times(0+1+2+\cdots+201)+8 \times 202=204626$.