The positive integer $k$ has the property that for all $m \in \mathbb{N}$, $k \mid m \implies k \mid m_r$. Show that $k \mid 99$. Question from Engel's book Problem solving strategies.
The positive integer $k$ has the property that for all $m \in \mathbb{N}$, $k \mid m \implies k \mid m_r$, where $m_r$ is the reflection of $m$, i.e. if $m=1234$ then $m_r = 4321$. Show that $k \mid 99$.
I start with a small case, say $k$ divides some 2 digit number $ab$. Then $k$ divides $ba$ also. Since $ab = 10a+b$ and $ba=10b+a$, I eliminate $a$ to get $k \mid -99b$. $b$ is a 1 digit number, so I think I need to use this fact somehow, but I am stuck here.
For the 3 digit case, let the number be $abc$. By similar reasoning, I get that $k$ divides $99a-99b = 99(a-b)$. Again, a-b is small, so maybe I can brute force this.
I have considered trying to show that $k$ must be a palindrome? Since $k \mid k$, we have $k \mid k_r$, and maybe try to get something? Thanks for the help!
 A: Let's just look at the divisibility rules for 3, 9, and 11. If 3 or 9 divides a number's sum of its digits then it's divides the number. If 11 divides the number you get from the alternating sum of its digits, then 11 divides the number.
Let's enumerate the digits of $m$ with subscripts so that $m_1$ is the first digit and $m_n$ is the last where $m$ is an $n$ digit long number.
The sum of the digits is $S=\sum_{i=1} ^n m_i$, and the alternating sum $S_a=\sum_{i=1} ^n (-1)^{n+1}m_i$. 
If $\ 9 \ | \ S$, then $ \ 9\  | \ m_r \ $ just by the law of communicative addition.  
Similarly if $\ 11\ |\ S_a$, then $ \ 11 \ | \ m_r $ .  The reason being that if $n$ is odd the same digits will be negative in the sum $S_a$ regardless of the order being reversed, and if $n$ is even then the sum will be opposite in parity but the same magnitude.
Ex:  Odd $n$: the number $94572$
$m_r: 94572 \to 27549$
$9-4+5-7+2=2-7+5-4+9$
Ex:  Even $n$:the number $9457$
$m_r: 9457 \to 7549$
$9-4+5-7 = - (7-5+4-9)$
A: HINT
Write $$k = \sum_{i=0}^{r_k}10^ix_i$$
where $x_i \in \{0,...,9\}$.
Since $k | k \implies k | k_r$, we have
$$\sum_{i=0}^{r_k}10^ix_i \quad \biggr{|} \quad\sum_{i=0}^{r_k}10^ix_{(r_k-i)},$$
$$\implies \lambda\sum_{i=0}^{r_k}10^ix_i = \sum_{i=0}^{r_k}10^ix_{(r_k-i)},$$
$$\implies 0 = \sum_{i=0}^{r_k}10^i(x_{(r_k-i)}-\lambda x_i),$$
$$\implies 0 = x_{(r_k-i)}-\lambda x_i \quad \forall i.$$
We therefore have that $k$ is a palindrome. 
