2013 AMC 10B Problem 18

The numbers run from $1001$ to $2012$. Split them into the ones starting with $1$ and the ones starting with $2$.

Numbers from $1001$ to $1999$.

1. Write such a number with digits $1, b, c, d$. We need $d = 1 + b + c$.
2. Once $b$ and $c$ are chosen, $d$ is fixed. The only catch is that $d$ must be a single digit, at most $9$, so $b + c$ can be at most $8$.
3. Count the pairs $(b, c)$ for each possible sum:

(For example, sum $2$ has the pairs $(0,2), (1,1), (2,0)$.) The total is $1 + 2 + \cdots + 9 = 45$.
4. The pair $b = c = 0$ gives $1001$, which is more than $1000$, so it counts.

Numbers from $2000$ to $2012$.

1. These have digits $2, 0, c, d$ with $c = 0$ or $1$. We need $d = 2 + c$. With $c = 0$ we get $2002$, which works. With $c = 1$ we get $2013$, which is not less than $2013$, so it is out. That is $1$ more number.

Total: $45 + 1 = 46$.

The answer is $\boxed{\textbf{(D)}\ 46}$.

Once the first digit is fixed, the last digit is decided by the middle two, so the problem becomes counting pairs of digits with a limited sum, which gives $1 + 2 + \cdots + 9$. The upper limit $2013$ is a deliberate edge: always check the short, awkward range at the end separately.

Write the number as $\overline{abcd}$ with $a \in \{1, 2\}$ and the condition $d = a + b + c$, which requires $a + b + c \le 9$.

Thousands digit $1$. Numbers $1001$ to $1999$: we need $d = 1 + b + c$, so $b + c \le 8$, and each such pair $(b, c)$ determines $d$ uniquely. For each sum $s = b + c$ from $0$ to $8$ there are $s + 1$ pairs, so the count is
$$1 + 2 + \cdots + 9 = 45.$$
(The pair $b = c = 0$ gives $1001$, which is greater than $1000$ and allowed.)

Thousands digit $2$. Only $2000$ to $2012$ are in range, so $b = 0$ and $c \in \{0, 1\}$. With $c = 0$, $d = 2$: $2002$ works. With $c = 1$, $d = 3$ gives $2013$, which is excluded. That is $1$ more.

Total: $45 + 1 = 46$.

The answer is $\boxed{\textbf{(D)}\ 46}$.