Octal divisibility

In base \(10\), there are simple divisibility rules formulated for division by \(2,3,4,5,6,8,9\) and \(11\) based on last digit, sum of digits or difference of alternate digits,(i.e., all numbers less than or equal to \((10+1)\). The number \(1\) is of course ignored in this context.

Which is the only number less than or equal to \(9_{10}=11_{8}\), for which a divisibility rule cannot be formulated in base \(8\), based on the either last digit(s), sum of digits or difference of alternate digit sums. The answer is to be given in base \(10\)?.

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