How to find the count of d digit numbers such that N is divisible by Rev(N)?

Question

How to find the count of d digit numbers such that N is divisible by Rev(N)?

For example if d = 4, then answer = 2

Since There are only 2 such numbers 8712 and 9801

8712 % 2178 = 0

9801 % 1089 = 0

Answer 1

From some searching, one comes upon https://oeis.org/A008919 as well as https://oeis.org/A031877 and gives a name to the numbers you are interested in, Palintiples.

Reading the information provided shows that this is a known problem with a solution. The number of (non-trivial) $n$-digit palintiples will apparently be:

$$2\cdot F\left(\left\lfloor \dfrac{n-2}{2}\right\rfloor\right)$$

where $F(k)$ is the $k$'th Fibonacci number.

To prove this, one must first accept or prove the claim that all such digit reversible numbers are precisely those numbers who are created by "symmetric concatenation" of blocks of $8712$ or $9801$ (not both) with $9$'s inserted in the middle of these blocks and/or $9$'s or $0$'s inserted between the copies of these in a symmetrical way. (Note: unlike $9$'s, the $0$'s may not be inserted within a block, only between blocks)

For example, $\color{blue}{8712}8712$, we can insert some $9$'s in the middle of the blue copy which we then to maintain symmetry insert near the end as well as $\color{blue}{87}\color{red}{99}\color{blue}{12}87\color{red}{99}12$. We can further insert some $0$'s between the two copies in the very middle as $879912\color{red}{00}879912$ and this will be a palintiple.

This becomes a question then of whether the base building block was $8712$ or $9801$ (hence the multiple of $2$ out front), followed by the quantity and positions of the inserted $9$'s and $0$'s.