Trailing zeroes in product How many trailing zeroes can be found at the end of $1!*2!*3!*...*100!$ ?
In each of the factors, the first $0$ is found in $5!$ where $2*5=10$. A second one is added in $10!$, a third one in $15!$ etc. 
So we start with $5$ factors with one $0$, then another $5$ factors with two $0$'s, and so on.
But how do we find the total number? Is it 5*1+5*2+5*3+...+5*24+1?
 A: It suffices to count the number $n$ of factors $5$ (the number of factors $2$ is greater than $n$). Note that 
$$1!\cdot 2!\cdot 3!\cdots 100!=\prod_{k=2}^{100}k^{101-k}$$
Hence, since $25$ is the largest power of $5$ which is less or equal to 100,
$$n=\sum_{j=1}^{20}(101-5j)+\sum_{j=1}^{4}(101-25j)=24\cdot 101-5\binom{21}{2}-25\binom{5}{2}=1124.$$
A: You are right to be looking at the factors of $5$, since there will only be more factors of two.
To find the number, notice that all factorials at least $5!$ have $5$ in them. All factorials larger than $10!$ have another $5$ in them. This, naively, gives
$$96 + 91 + \dots + 1$$
But we're not done. Now all factorials at least $25!$ have $25$ in them, in which case we need to add another factor of $5$. Similarly all factorials at least $50!$ have $25$ in them, and so forth. This gives
$$76 + 51 + 26 + 1$$
and now we're done because $125 > 100$. This gives a final answer of $1124$.

Looking from a more general perspective, it is well-known and easy to see that for each prime $p$, the number of factors of $p$ in $n!$ is
$$\sum_{k=1}^{\infty} k\lfloor \frac{n}{p^k} \rfloor.$$
So then we need to take the sum
$$\sum_{n = 1}^{100} \sum_{k=1}^{\infty} k\lfloor \frac{n}{5^k} \rfloor.$$
We were just taking this sum "horizontally" for the $k=1$ term and then the $k=2$ term, noticing that it vanishes for $k > 2$.
