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 \bmod i~ 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 ~10\,007~ after every iteration.

The final answer is the sum of the above counts for all of ~F_1~'s divisors, again taken modulo ~10\,007~. The time complexity of this algorithm is ~\mathcal O(\sqrt{F_1} \times N)~.