Romanian Mathematical Olympiad

The numbers $\overline{abc},\overline{bca},\overline{cab}$ have the same sum of digits therefore they will have the same remainder $r$ when divided by 9. Since the remainders modulo 27 are small, they are preserved modulo 9. Indeed if $n=27k+r$, then $n=9\cdot 3k+r$, so $r$ will be a common remainder. 27 must divide the difference $\overline{bca}-\overline{abc}=90b+9c-99a=9(a+b+c)+81b-108a=9(a+b+c)+27(3b-4a)$, therefore $3\mid a+b+c$ which implies $r=3$.

The numbers $abc$, with $a<b<c$ and $a+b+c=12$ are 129, 138, 147, 156, 237, 246 and 345. Only 138 and 246 have the remainder 3 modulo 27. The numbers $abc$, with $a<b<c$ and $a+b+c=21$ are 489, 579 and 678. Convenient are 489 and 678. In the end we have 4 solutions: 138, 246, 489 and 678.