Giả sử 0<a<b<c. Theo đề bài
\(\overline{abc}+\overline{acb}=200a+11b+11c=499\)
\(\Rightarrow11\left(a+b+c\right)=499-189a=495+4-187a-2a\)
\(\Rightarrow11\left(a+b+c\right)=45.11-17.11.a+\left(4-2a\right)\)
\(11\left(a+b+c\right)⋮11\Rightarrow145.11+17.11.a+4-2a⋮11\)
\(\Rightarrow4-2a⋮11\Rightarrow a=2\) Thay a=2 vào biểu thức
\(11\left(a+b+c\right)=499-189a\Rightarrow a+b+c=11\)