Let's break down our counting of valid recipes by the number of monkeys $m$ which would be served by them. Note that $m$ uniquely determines the value of $R_1$ – in particular, to satisfy $m\times R_1=F_1$, we must have $R_1=\frac{F_1}{m}$. This further implies that $m$ must be a divisor of $F_1$.

This is useful due to the fact that $F_1$ can't have many different divisors. In fact, we can find all of them by iterating over each integer $i$ between $1$ and $\lfloor \sqrt{F_1}\rfloor$, inclusive. If $F_{1}modi$ is equal to $0$, then $i$ is a divisor of $F_1$. Furthermore, $\frac{F_1}{i}$ is also a divisor of $i$ – however, if $i=\frac{F_1}{i}$, we need to be careful to not count its recipes twice!

Now, for a given value of $m$, we need to be able to count the number of valid recipes which would serve $m$ monkeys. As shown, there's a single valid value for $R_1$. Meanwhile, each other $R_i$ (for $2\le i\le N$) can independently be any positive integer such that $m\times R_i\le F_i$. The number of such valid values is simply equal to $\lfloor \frac{F_i}{m}\rfloor$. Overall, the number of valid recipes for a given value of $m$ is then the product of these $N-1$ rounded-down quotients. While computing the running product, we need to be careful to mod it by $10007$ after every iteration.

The final answer is the sum of the above counts for all of $F_1$'s divisors, again taken modulo $10007$. The time complexity of this algorithm is $\mathcal{O}(\sqrt{F_1}\times N)$.