18. How many zeros are there at the end of $\frac{2018!}{30!\times 11!}$ ?
Solution
493

Each zero at the end of $N$ ! must be generated by a $(2 \times 5)$ in its prime factorisation. In general $N!=2^{a} \times 5^{b} \times K$. For all values of $N, a$ will be greater than $b$. We therefore determine the value of $b$ in each of 2018!, 30! and 11!. In 2018! $=1 \times 2 \times \ldots \times 2018$ there are
- 403 multiples of five;
- 80 multiples of $5^{2}=25$, each contributing one additional five;
- 16 multiples of $5^{3}=125$, each contributing one further additional five; and
- 3 multiples of $5^{4}=625$, each contributing one yet further additional five.

Therefore, for 2018!, $b=403+80+16+3=502$.
In 30 ! there are 6 multiples of five and 1 multiple of 25 . So, for 30 !, $b=6+1=7$.
In 11 ! there are 2 multiples of five. So, for $11!, b=2$.
Therefore the power of five in the fraction $\frac{2018!}{30!\times 11!}$ is $502-7-2=493$ and so there are 493 zeros at the end of the number.