2013 AMC 10B 第 18 题

Count numbers from $1001$ to $1999$ by their units digit

Handle the small range from $2000$ to $2012$ separately

Once the first three digits are given, their sum determines the units digit, provided that the sum is at most $9.$

We separate the possibilities according to the thousands digit.

If the thousands digit is $1,$ then the sum of the hundreds and tens digits must be at most $8.$ For each possible sum $s,$ there are $s+1$ choices, because the hundreds digit can range from $0$ to $s.$

Thus, this case gives the ninth triangular number: $$\dfrac{9\cdot 10}2 =45.$$

If the thousands digit is $2,$ the only qualifying number below $2013$ is $2002,$ giving $46$ cases in total.

Thus, the correct answer is D.