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!

The division test for $3$ and $9$ (i.e if a number's digits sum to a multiple of $3$, than the number is a multiple of $3$, and similarly for $9$), shows that the reversal of any multiple of these numbers is also a multiple of these numbers. 
Obvious restrictions on the first digit gives conditions for multiples of $2$ and $5$. This also gives a condition for $6$, since any multiple of $2$ and $3$ is a multiple of $6$. For $4$, we just require that the first two digits (when reversed) are divisible by $4$. A divisibility test for $8$ is to see if the last three digits of a number are divisible by $8$. This can be used to give a condition for $8$. For $7$, a restriction could be placed, but it seems a bit more complicated. 
I'm not sure what the algorithm question is getting at (for instance, just take any multiples of $3$. Does this satisfy your question?). 
