25. What is the largest number of digits that can be erased from the 1000 -digit number 201820182018.... 2018 so that the sum of the remaining digits is 2018 ?
A 343
B 582
C 671
D 741
E 746
25. D Since we want as few digits remaining in the number as possible, we require as many $8 \mathrm{~s}$ (the highest digit) as possible. Also, since $2018=252 \times 8+2$ and there are only $2508 \mathrm{~s}$ in the 1000 -digit number, some $2 \mathrm{~s}$ (the next highest digit) will also be required. Therefore, since $2018-250 \times 8=18$ and $18=9 \times 2$, the smallest number of digits remaining for the sum of these digits to be 2018 is $250+9=259$. Therefore the largest number of digits that can be erased is $1000-259=741$.