Sum of repeating decimals

Compute the sum of all numbers of the form $0.\overline{abc}$ (i.e. $0.abcabcabc\dots$ ) where $a, b, c$ are distinct digits from $0$ to $9$ .

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$0.\overline{abc}=(100a+10b+c)/999$ . Sum over the $10\cdot9\cdot8$ distinct-digit triples.
By symmetry each position sums to $72\cdot45=3240$ ; total $=111\cdot3240/999$ .

Answer

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360

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