2013 AMC 10B Problem 18

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.