2013 AMC 10B Problem 18

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 $\boxed{\textbf{(D)}\ 46}$.

Once the leading digit is fixed, the units digit is determined by the middle two, so the problem is counting pairs $(b, c)$ with a bounded sum: a triangular number. The upper limit $2013$ is a deliberate edge: it forces a separate check of the short $2000$s range.

• Forgetting the 2000-2012 range (answering 45), or allowing b + c = 9, which would make the units digit 10.
• Including $2013$ itself, or counting $2011$ (units digit $1 \ne 3$) as valid.