Given a three-digit positive integer $A$, swap the last two digits and add the resulting number to the given number, thus obtaining a four-digit number starting with 173. Find the number $A$.

Swap the last two digits of the given three-digit positive integer $A$ and add the resulting number to the given number, thus obtaining a four-digit number starting with 173. Find the number $A$.
Let $\overline{\alpha \beta \gamma}$ be the given three-digit number, then the new three-digit number is $\overline{\alpha \gamma \beta}$. Therefore, we have
$$
\begin{array}{l}
\overline{\alpha \beta \gamma}+\overline{\alpha \gamma \beta}=100 \alpha+10 \beta+\gamma+100 \alpha+10 \gamma+\beta \\
=200 \alpha+11(\beta+\gamma) .
\end{array}
$$

Since this is a four-digit number starting with 173, we have $\alpha=8$.
Let $x$ be its last digit, then we have
$$
1600+11(\beta+\gamma)=1730+x,
$$

which simplifies to $11(\beta+\gamma)=130+x$.
From this relationship and 111132, we get $x=2$.
Thus, the four-digit number is 1732.
Therefore, $\beta+\gamma=12$ and $\overline{8 \beta \gamma}+\overline{8 \gamma \beta}=1732$.
Hence, the given three-digit positive integer $A$ is one of $839,848,857$,
$$
866,875,884,893 \text { . }
$$