Proof: if $121 \mid x$ and $121 \mid \operatorname{rev}(x)$ then digitsum($x$) is even $\newcommand{\rev}{\operatorname{rev}}$I ran a python script and whenever $121 \mid x$ and $121 \mid \rev(x)$ then $\operatorname{digitsum}(x)$ was even. Can someone prove that this is always true?
$\rev(x)$ is the reverse of $x$, e.g. $\rev(1234) = 4321$
and digitsum is the sum of the digits of a number, e.g. $\operatorname{digitsum}(1234) = 10.$
 A: A counterexample is the $45$-digit palindrome
$$
705050505050505050505070505050505050505050507,
$$
whose digit sum is $121$.
(Hints for its construction: the multiplicative order of $10$ modulo $121$ equals $22$. So palindromes of length $22k+1$ have the property that each pair of symmetric-around-the-centre digits $d$ contributes either $2d$ or $-2d$ to the number's value modulo $121$.)
A: We work first in the ring $R=\Bbb Z/2\cdot11^2=\Bbb/242$.
The multiplicative order of the unit $10+121=131\in R$ is $22$.
Its inverse is $109$.
Now we are searching for digits $a_0,a_1,a_2,\dots,a_N$ for some suitable integer $N>0$ so that the following conditions are satisfied in $R$:
$$
\begin{aligned}
a_0+a_1\cdot 131+a_2\cdot 131^2+\dots &\in\{0, 121\}\ ,\\
a_0+a_1\cdot 109+a_2\cdot 109^2+\dots &\in\{0, 121\}\ ,\\
a_0+a_1+a_2+\dots &\in\{1, 3, 5, \dots, 119, 121, 123, \dots, 243\}\ .
\end{aligned}
$$
The first equality to $0$ or $121$ of the LHS, taken modulo $121$, is equivalent to the fact that the decimal number $\overline{a_N\dots a_2a_1a_0}$ is divisible by $121$.
The second relation, taken modulo $121$, (and multiplied with the corresponding power of ten) is equivalent to the fact that the number $\overline{a_0a_1a_2\dots a_N}$ is divisible by $121$.
The last equality wants an odd sum of digits.
If we pass in any of the first two relation modulo two we see that only $121$ can be obtained. So it is enough (and necessary) to insure modulo $242$:
$$
\begin{aligned}
a_0+a_1\cdot 131+a_2\cdot 131^2+\dots &=121\ ,\\
a_0+a_1\cdot 109+a_2\cdot 109^2+\dots &=121\ .
\end{aligned}
$$
(We have some explanation for the fact that solutions are "rather big".)
Let us solve these equations, first forgetting about the "digit character" of $a_0,a_1,a_2,\dots, a_N$. We may use only $a_0=121$ and that's all. However, this is not a decimal digit, so we need to force it "from the pieces". We consider the integer numbers $10^0$, $10^{22}$, $10^{44}$, and so on as the "pieces". We can take them all with the digit one, so that the number we obtain
$$
S=1+10^{22}+10^{44}+\dots+10^{22\cdot 120} 
$$
is "nice". Or we can use the digit $9$ as often we can to have a "small" number. For instance
$$
T = 9(1+10^{22}+10^{44}+\dots+10^{22\cdot 12})+4\cdot 10^{22\cdot 13}\ .
$$
(Feel free to move the digit $4$ instead of one of the nines.)

We can further search for "smaller solutions", but these ones above, based on $10^{22}$ are simple to write and simple to understand.
