This is the formula we use to calculate the number of trailing zeros in $n!$ but we have to add the individual elements of the sum to get up with an answer. Is there any way we can bypass that?

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    $\begingroup$ For a generic $n$, I think that there is no magic formula... $\endgroup$ – Robert Z Oct 11 '16 at 14:03

Let $r=1/5$.

Without the floor function, we have


With the floor function, we have

$$\sum_{i=1}^k\left\lfloor nr^i\right\rfloor\le\left\lfloor n\frac{r-r^{k+1}}{1-r}\right\rfloor$$

We also have that for $k\ge a=\lceil\log_{1/r}(n)\rceil:$

$$\sum_{i=1}^k\left\lfloor nr^i\right\rfloor=\sum_{i=1}^a\left\lfloor nr^i\right\rfloor\le\left\lfloor n\frac{r-r^{a+1}}{1-r}\right\rfloor$$

which seems to be pretty close to the correct value.enter image description here


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