What multiples of $d$ are still multiples of $d$ when they have their digits reversed? I teach at a school for 11 to 18 year olds. Every term I put up a Challenge on the wall outside my classroom.
This question is one that I have devised for that audience. I think that it is quite interesting and I would like to share it with a wider audience. I am not aware of it being a copy of another question.
An underlying assumption to this question is that we are writing numbers in base 10. A further question could be to investigate the same question in other bases.
Define the reverse of a number as the number created by writing the digits of the original number in the opposite order. Leading zeroes are ignored, although you will see that this makes very little difference to my question if this stipulation is removed.
Thus $Reverse(1456)=6541$ and $Reverse(2100)=12$
My question is this: For a given value $d$, what numbers $x$ have the property that $x$ is a multiple of $d$ and $Reverse(x)$ is a multiple of $d$?
I call such numbers "reversible multiples of $d$."
Clearly there are "brute force" ways to investigate this question, but I am looking for more subtle answers.
A good answer should address the following:
a) For some values of $d$ all multiples are reversible multiples. List those values, with a proof of why this is so.
b) For other values there are certain properties that the multiple must have for it to be a reversible multiple. Explain these.
c) For at least one value there is an algorithm that can be used to construct reversible multiples. Describe such an algorithm.
Enjoy!

Interesting question!  I see no obvious route toward a general solution yet.
For $d = 2^k$, the solution is not very interesting: Numbers that qualify are those whose last $k$ digits are divisible by $d$, and whose first $k$ digits, when reversed, are also divisible by $d$.
For $d = 3$ or $9$, any number divisible by $d$ remains divisible by $d$ when reversed.
For $d = 5$, the solution is again not very interesting: The number must begin with 5, and end in either 0 or 5.
For $d = 6$, any number divisible by $6$ that begins with an even digit remains divisible by $6$ when reversed.
There are some interesting divisibility tests for $d = 7$, but I'm still working on how to turn them into conditions for divisibility surviving digit reversal.  I'll come back to edit this if/when I come up with something.
There are, of course, no numbers divisible by $10$ that remain so when reversed.  (Note that this is affected by whether leading zeros are dropped.)
Any number divisible by $11$ remains so when reversed.
Analogously with the $d = 6$ case, any number divisible by $12$, that begins with two digits that yield a number divisible by $4$ when reversed, remains divisible by $12$ when reversed.
