2017 AMC 10B Problem 16

For numbers less than $100,$ we only have a $0$ if it is a multiple of $10,$ of which there are $9.$ For numbers between $100$ and $999$ inclusive, we will use complementary counting. There are $900$ total numbers in this range. Also, there are $9\cdot 9\cdot 9=729$ numbers in this range with no $0$ since there are $9$ ways to choose each digit to not be $0.$ Thus, the total in this range is $171.$ For numbers between $1000$ and $1999$ inclusive, we will use complementary counting again. There are $1000$ total numbers in this range. Also, there are $1\cdot 9\cdot 9\cdot 9=729$ numbers in this range with no $0$ since there are $9$ ways to choose each of the last $3$ digits to not be $0$ and the first digit must be $1.$ Thus, the total in this range is $271.$ There are $18$ numbers between $2000$ and $2017$ inclusive, each with a $0$ in the second digit from the left. This makes the total $$9+171+271+18=469.$$ Thus, the correct answer is A .