An integer is divisible by nine precisely when the sum of its decimal digits is divisible by nine

Prove that if a = \sum_{i=0}^n a_i 10^i is any positive integer then a = \sum_{i=0}^n a_i\ (\mod 9). Note that this is the usual arithmetic rule that the remained after division by 9 is the same as the sum of the decimal digits mod 9 – in particular an integer is divisible by 9 if and only if the sum of its digits is divisible by 9.)

Letting bars denote arithmetic mod 9, we have \overline{a} = \overline{\sum_{i=0}^n a_i 10^i} = \sum_{i=0}^n \overline{a_i} \overline{10}^i = \sum_{i=0}^n \overline{a_i} \cdot 1, and we’re done.

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