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How do I proof that the sum of the digits of these two numbers:

$10^n-T$ and $10^{n+1}-T$ where n and T are positive integers, are not at the same time odd or even, i.e. if one of the sums is odd then the other is even and vice versa.

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    $\begingroup$ Consider $n=1$ and $T=33$ $\endgroup$ – Henry Nov 17 '14 at 11:35
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(Assuming both expressions and $T$ are positive) $$(10^{n+1}-T)- (10^n-T) = 9\times 10^n,$$ i.e. they have the same digits except that the larger one has an extra $9$.

So the sums of their digits differ by $9$.

$9$ is odd.

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