Five, a chicken farm uses chicken coops to hold chicks. If each coop holds 36 chicks, then 11 chicks are left over. If two fewer coops are used, then all the chicks can be evenly distributed. It is known that one coop can hold a maximum of 45 chicks. How many chicken coops were there originally? How many chicks are there?
Five, Solution: Let the original number of chicken coops be $x$, and when two coops are reduced, each coop holds $n$ chicks. Then $x \geqslant 3$, $n \leqslant 45$, and $36 x+11=n(x-2)$. Therefore, $n=36+\frac{83}{x-2}$.

Since $n$ is an integer, it follows that $x-2$ must divide 83, and since 83 is a prime number, $x-2=1$ or $x-2=83$. Thus, $x=3$ or 85.

When $x=3$, $n=119>45$, which does not meet the condition, so it is discarded. When $x=85$, $n=37$.

Therefore, the original number of chicken coops is 85, and the number of chicks is $83 \times 37 = 3071$.