Sum of the digits of two numbers

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.

• Consider $n=1$ and $T=33$ – Henry Nov 17 '14 at 11:35

(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.