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that every natural number minus its palindrome is divisble by 9. Let \(\sum_{i=1}^{n}10^{i-1}a_i = a, a_i \in {0,...,9}\).
Its palindrome is \(\sum_{i=1}^{n}10^{n-(i-1)}a_i = \overline{a}\).
Thus \(a - \overline{a} = \sum_{i=1}^{n}a_i(10^{i-1} - 10^{n-(i-1)})\), or we know that \(\forall a,b \in \mathbb{N}, 10^a - 10^b \equiv 0\pmod 9\).
Consequently, \(a - \overline{a} \equiv 0\pmod 9\).

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