# 2. Option 1.

In a certain three-digit number $N$, the last two digits were swapped and the resulting number was added to the original. The sum turned out to be a four-digit number starting with 173. Find the largest possible sum of the digits of the number $N$.
Answer: 20.

Solution: Let the original number be $\overline{a b c}$, and the last digit of the sum $\overline{a b c}+\overline{a c b}$ be $x$. Then, $100 a+10 b+c+100 a+10 c+b=200 a+11(c+b)=1730+x$. If $a1730+x$.

Therefore, $a=8$. Thus, $(800+10 b+c)+(800+10 c+b)=1730+x$, i.e., $11(b+c)=130+x$. From this, it follows that $x=2$ (among the numbers from 130 to 139, only 132 is divisible by 11). Therefore, $b+c=132 / 11=12$. Answer: $a+b+c=8+12=20$.