Problem 1296

1. Let the original 3-digit number be represented as $\overline{abc}$, where $a$, $b$, and $c$ are its digits. This can be expressed as $100a + 10b + c$.
2. When we swap the last two digits, the number becomes $\overline{acb}$, which can be expressed as $100a + 10c + b$.
3. According to the problem, the sum of these two numbers is a 4-digit number starting with 173. Therefore, we have:
$$\overline{abc} + \overline{acb} = (100a + 10b + c) + (100a + 10c + b) = 200a + 11b + 11c$$
4. The resulting number is a 4-digit number starting with 173, so:
$$1730 \leq 200a + 11(b + c) \leq 1739$$
5. To find the value of $a$, we solve for $a$:
$$1730 \leq 200a + 11(b + c) \leq 1739$$
Dividing the entire inequality by 200:
$$8.65 \leq a + \frac{11(b + c)}{200} \leq 8.695$$
Since $a$ must be an integer, the only possible value for $a$ is 8.
6. Substituting $a = 8$ back into the inequality:
$$1730 \leq 1600 + 11(b + c) \leq 1739$$
Simplifying:
$$130 \leq 11(b + c) \leq 139$$
7. Dividing the entire inequality by 11:
$$11.818 \leq b + c \leq 12.636$$
Since $b + c$ must be an integer, the only possible value for $b + c$ is 12.
8. Therefore, the digits $b$ and $c$ must sum to 12. There are several pairs of digits $(b, c)$ that satisfy this condition, such as $(3, 9)$, $(4, 8)$, $(5, 7)$, $(6, 6)$, $(7, 5)$, $(8, 4)$, and $(9, 3)$.

The final answer is $\boxed{1732}$.