Find the prime factor decomposition of $100!$ and determine how many zeros terminates the representation of that number. Find the prime factor decomposition of $100!$ and determine how many zeros terminates the representation of that number.


Actually, I know a way to solve this, but even if it is very large and cumbersome, and would like to know if you have an easier way, or if I am applying wrong.
Setting by $\left[\frac{b}a \right]$ the quotient of $b$ with $a$, we have too $E_p(m)$ the largest exponent power $p$ dividing $m$, and found the demonstration of a theorem that says (that in my text says it was discovered that Lagendre) $$E_p(n!)=\left[\frac{n}p \right]+\left[\frac{n}{p^2} \right]+\left[\frac{n}{p^3} \right]+\;...$$always remembering that there will be a number $s$ such that $p^s\geq n!$ which tells us that $$\left[\frac{n!}{p^s} \right]=0$$ thus making the sum of a finite $E_p(n!)$. So that I can address the first question I asked, really have to get all the cousins $(p_1,p_2,...,p_k)$ and make all $$E_{p_1},\;E_{p_2},\;E_{p_3},\;...,\;E_{p_k}$$ with $p$ and cousin $1&lt;p&lt;100$. And to find the zeros have to see how we both exponents in numbers 5 and 2.


  Example $$10!=2^83^45^27\\p&lt;10\\E_2(10!)=5+2+1=8\\E_3(10!)=3+1=4\\E_5(10!)=2\\E_7(10!)=1\\$$
It's possible to demonstrate that if N is a multiple of 100, N! ends with (N/4)-1 zeroes.
It's possible to demonstrate that if N is a multiple of 100, N! ends with (N/4)-1 zeroes.