10
$\begingroup$

I know the formula to calculate this, but I don't understand the reasoning behind it:

For example, the number of trailing zeros in $100!$ in base $16$:

$16=2^4$,

We have: $\frac{100}{2}+\frac{100}{4}+\frac{100}{8}+\frac{100}{16}+\frac{100}{32}+\frac{100}{64}=97$

Number of trailing zeros$ =\frac{97}{4} = 24$.

Why do we divide by the power of '$2$' at the end?

  • 1
    For some basic information about writing math at this site see e.g. [here](http://meta.math.stackexchange.com/questions/5020/), [here](http://meta.stackexchange.com/a/70559/155238), [here](http://meta.math.stackexchange.com/questions/1773/) and [here](http://math.stackexchange.com/editing-help#latex).2012-11-01
  • 0
    I'm pretty sure this is a dupe of an earlier question on using de Polignac-Legendre...2012-11-01

3 Answers 3