Reversal of Number Multiplied by 11

Theorem
Let $n \in \N$ be a number for which, when written in decimal notation, no two adjacent digits total to more than $9$.

Let $n'$ denote the reversal of $n$.

Then $n \times 11$ is the reversal of $n' \times 11$.

Proof
By Basis Representation Theorem, there exists one and only one sequence $\sequence {r_j}_{0 \mathop \le j \mathop \le t}$ such that:


 * $(1): \quad \ds n = \sum_{k \mathop = 0}^t r_k 10^k$
 * $(2): \quad \ds \forall k \in \closedint 0 t: r_k \in \N_{10}$
 * $(3): \quad r_t \ne 0$

Since no two adjacent digits of $n$ total to more than $9$, we have:
 * $\paren {r_i + r_{i - 1} } \in \N_{10}$ for $i = 1, 2, \dots, t$

Now:

Since $r_0, r_t$ and each $r_k + r_{k - 1}$ is in $\N_{10}$, the above is the unique representation of $n \times 11$.

The reversal of $n$, $n'$, is given by:
 * $\ds n = \sum_{k \mathop = 0}^t r_{t - k} 10^k$

Hence:

Since $r_t, r_0$ and each $r_{t - k} + r_{t - k + 1}$ is in $\N_{10}$, the above is the unique representation of $n' \times 11$.

Its reversal is given by:
 * $\ds r_0 + \paren {r_0 + r_1} 10^1 + \paren {r_1 + r_2} 10^2 + \dots + \paren {r_{t - 1} + r_t} 10^t + r_t 10^{t + 1}$

which is:
 * $\ds r_0 + \sum_{k \mathop = 1}^t \paren {r_{k - 1} + r_k} 10^k + r_t 10^{t + 1}$

which can be seen to be equal to $n \times 11$.