If $N$ is a multiple of $100$, $N!$ ends with $\left(\frac{N}4-1 \right)$ zeroes.

Question

Did certain questions about factorials, and one of them got a reply very interesting that someone told me that it is possible to show that

If $N$ is a multiple of $100$, $N!$ ends with $\left(\frac{N}4-1 \right)$ zeroes.

Sought such a demonstration, not found when trying to do there, I thought of induction, but did not fit, I thought of several ways, but did not get any progress, would like to know how to do, or to see a demonstration of this.

I thought about trying to calculate the following series $$\sum_{k=0}^{\infty} \left[\frac{n}{5^k}\right]$$ only that I was in doubt because it is using the quotients $\left[\frac{n}{5^k}\right]$

Answer 1

Hint:

A number ending with zero is divisible by $10=2\times5$. Hence we have to find the multiplicities of $2$ and $5$. However the multiplicity of $2$ clearly dominates that of $5$. Hence it suffices to find the multiplicity of $5$.

Number of multiples of $5$ = $\left\lfloor\frac{N}{5}\right\rfloor$

Number of multiples of $25$ = $\left\lfloor\frac{N}{5^2}\right\rfloor$

Number of multiples of $125$ = $\left\lfloor\frac{N}{5^3}\right\rfloor$ ...

Answer 2

It's not true. Consider $N = 600$.

$$\left\lfloor\frac{600}{5} \right\rfloor + \left\lfloor \frac{600}{25}\right\rfloor + \left\lfloor \frac{600}{125}\right\rfloor = 120 + 24 + 4 = 148 = \frac{600}{4} - 2.$$