Is there a formula that verifys if a number is a palindromic number? Let's say I need to verify that a very long number is a palindromic number. Is there a formula that can be used to determine if it is one?
 A: We could formalize your problem as the following. If $n \in \mathbb{N} $ is palindromic, then it is true, that
$$n = r(n),$$
where $r : \mathbb{N} \rightarrow \mathbb{N}$ function gives the reverse of a number.
We can get $r$ like this:
$$r(n)=\sum_{\ell=0}^{m-1}10^{m-1-\ell}\left(\left\lfloor\frac{n}{10^\ell}\right\rfloor-10\left\lfloor\frac1{10}\left\lfloor\frac{n}{10^\ell}\right\rfloor\right\rfloor\right),$$
where $n$ is the number we want to reverse, and $m$ is the digits of $n$.
We know that $m$ is
$$m=\lfloor\log_{10}n\rfloor+1.$$
So using this we get for $r$ the following.
$$r(n)=\sum_{\ell=0}^{\lfloor\log_{10}n\rfloor}10^{\lfloor\log_{10}n\rfloor-\ell}\left(\left\lfloor\frac{n}{10^\ell}\right\rfloor-10\left\lfloor\frac1{10}\left\lfloor\frac{n}{10^\ell}\right\rfloor\right\rfloor\right),$$
where $n$ is the number we want to reverse.
After all you just have to compute $r(n)$ for a given $n$, and if $r(n)=n$ then $n$ was palindromic.
Of course there are much more easier solutions using some computer science.
