<s>[INST] 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 .
 [/INST] $1+16+16 \cdot 17+ 16 \cdot 17 \cdot 18 = 5185$. 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.
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