Problem 92

We take cases on the thousands digit, which must be either $1$ or $2$:
If the number is of the form $\overline{1bcd},$ where $b, c, d$ are digits, then we must have $d = 1 + b + c.$ Since $d \le 9,$ we must have $b + c \le 9 - 1 = 8.$ By casework on the value of $b$, we find that there are $1 + 2 + \dots + 9 = 45$ possible pairs $(b, c)$, and each pair uniquely determines the value of $d$, so we get $45$ numbers with the given property.
If the number is of the form $\overline{2bcd},$ then it must be one of the numbers $2000, 2001, \dots, 2012.$ Checking all these numbers, we find that only $2002$ has the given property.
Therefore, the number of numbers with the property is $45 + 1 = \boxed{\textbf{(D)}46}$ .