# Finding solutions of $\sum a_i = \prod a_i = n$ in $\{a \in \mathbb{Q} \mid 100a \in \mathbb{Z}^+\}$

Problem: Define the set $$Q_p := \{a \in \mathbb{Q} \mid 100a \in \mathbb{Z}^+ \}$$. Given an integer $$k$$ and some $$n \in Q_p$$, find $$a = \{a_1, a_2, \cdots a_k\} \subset Q_p$$ such that

$$\sum_{i=1}^k a_i = \prod_{i=1}^k a_i = n$$

I'll call $$Q_p$$ the "price rationals," as they're the fractions whose decimal form can be a price in dollars and cents. (Is there an actual standard form for a subset of this nature, i.e., the rationals that are all of the integers divided by $$m$$?)

This problem is a generalization of this week's Riddler Express. There, $$n = 7.11, k=4$$, and the solution is $$a = \{3.16,1.5,1.25,1.2\}$$. But I came to the answer via a bit of brute force and a bit of factorization ($$7.11 = 9\cdot 79 /100 \to 3.16 = 79/25$$), and I have to imagine there's a better way.

If $$k=2$$, then from Vieta's formulas it's clear that $$a_1, a_2$$ are the roots of $$x^2-nx+n = 0$$. Not all $$n$$ will have rational solutions, however; solutions exist only if $$n^2-4n = q^2$$ for some $$q^2 \in Q_p$$. So we have a method of determining whether there is a solution, and a method of finding it, for $$k=2$$. Great! (The same solution, with different existence parameters, would work if we allowed all positive rationals, not just the price rationals.)

But now consider $$k=3$$. If we try the same method, we get $$x^3 -nx^2 + (a_1a_2+a_1a_3+a_2a_3)x -n = 0$$. While two of our coefficients are equal to $$n$$, the third one... isn't. And there's no apparent rearrangement/transformation to get it in terms of $$n$$.

And it gets worse as $$k$$ increases, because we're really only given two equations, which is solvable in two variables, but not so much in three, four, etc. We can solve for $$a_2, a_3$$ in terms of $$n$$ and $$a_1$$ for $$k=3$$, but $$a_1$$ remains an unbound variable... except that it can't remain unbound, if there's a unique solution. (Which raises the question: can there be multiple solutions? EDIT: Yes! See answer by @paw88789)

Does anyone have thoughts on attacking this systematically for $$k=3,4$$, or even higher? Or do we end up with factorization followed by trial and error?

$$(0.80)(2.50)(3.30)=.80+2.50+3.30=6.60$$ and $$(1.10)(1.50)(4.00)=1.10+1.50+4.00=6.60$$
• Very nice, thank you for the check! At some point brute-forcing with SageMath is a path I'll probably go down, and knowing multiple solutions are possible is greatly useful! I suspect multiple solutions are more likely for abundant numbers (divided by $100$). Jun 29, 2022 at 22:45