The right-most non-zero digit

I could't work out any recursive formula or closed form for $n$ but here is the working process.

Define $x \in N$ such that $5^{x} \leq n < 5^{x + 1}$ .

$\forall a \in \mathbb N , a \leq x$ Find $\left\lfloor{\dfrac{n}{5^{a}}}\right\rfloor \pmod{10}$ , call it $b_a$

Assign the values as follows , call it $d_a$

$\begin{array}{l|r} b_a & d_a \\ \hline 1 & 1 \\ \hline 2 & 2 \\ \hline 3&1 \\ \hline 4& 4 \\ \hline 5 &4 \\ \hline 6&4 \\ \hline 7& 3 \\ \hline 8&4 \\ \hline 9&1 \\ \hline 0&1 \end{array}$

Evaluate : $\displaystyle\prod_{a = 1}^{a = x} d_a$ , call the value obtained to be $t$

Evaluate the number of trailing zeroes in $n!$ , say $m$ .

Evaluate $y \pmod{5}$ such that $2^{m}\times{y} \equiv{t}\pmod{5}$

The desired answer is $y + 5$

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For the given problem :

$x = 5 , t = 4 ,m = 2499 , y = 3 \Rightarrow \boxed{8}$ is the required answer.

just see and multiply ten numbers from 1 until 10, like as : 1.2.3.4.5.6.7.8.9.10 and the rightmost non-zero is 8