Zeros at the end of sum of factorials Find the number of zeros at the end of $15! + 16! + 17! + 18!$ ? 
I know the method find the number of zeros at the end of x! where $x = { 15! , 16! , 17! ...}$ by dividing by number by $5,5^2, 5^3$ and so on .
 A: $1+16+16 \cdot 17+ 16 \cdot 17 \cdot 18 = 5185$. If the first non-zero digit of $15!$ is odd, then there will be just $3$ zeroes at the end from the $15!$. However, if it is even, there will be exactly one more zero. (hint: look at integers from $1$ to $9$)
Then we have to find:
$$1 \cdot 2 \cdot 3 \cdot 4 \cdot 6 \cdot 7 \cdot 8 \cdot 9 \cdot 11 \cdot 12 \cdot 13 \cdot 14 \pmod {10}$$
and the result is an even number.
A: First divide $15!$ by $10^3$ (in the obvious way, from its expression) to determine its last nonzero digit. Hopefully, it will be the 4th digit from the right.
You're left with
$$N=3\cdot 4\cdot 6\cdot 7\cdot 8\cdot 9\cdot 11\cdot 12\cdot 13\cdot 7\cdot 3= 2^8\cdot 3^6\cdot 7^2\cdot 11\cdot 13$$
Modulo $100$, we have
$$N\equiv 56\cdot 29\cdot 49\cdot 43=56\cdot 29\cdot (50-1)(40+3)\equiv  56\cdot 29\cdot 7\equiv 56\cdot 3\equiv 68  $$
As a conclusion, $\; 15!\equiv 68\,000\mod 10^5$.
Now we'll compute:
\begin{align}M&=1+16+16\cdot 17+16\cdot 17\cdot 18=17^2+16\cdot 17\cdot 18=17(17+17^2-1)\\
&= 17(17+289-1)=17\cdot 305=5185
\end{align}
Thus, $$15!+16!+17!+18!\equiv68\,000\cdot 5185\equiv 8000\cdot 85=680\,000\equiv \color{red}{ 80000} \mod 10^5.$$
