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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?

Eric Snyder
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  • I doubt it's standard, but you could denote the set $Q_p$ by $\frac{{\mathbb Z}^+}{100}$ – paw88789 Jun 29 '22 at 13:58
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    You may get some ideas from Kurlandchik and Nowicki, When the sum equals the product, available at https://www-users.mat.umk.pl/~anow/ps-dvi/si-krl-a.pdf – see also https://math.stackexchange.com/questions/400778/sum-equal-to-the-product and https://math.stackexchange.com/questions/2919430/sum-equals-the-product and https://math.stackexchange.com/questions/286925/on-the-equality-of-the-sum-of-a-set-and-the-product-of-a-set and https://math.stackexchange.com/questions/1635875/find-a-thousand-natural-numbers-such-that-their-sum-equals-their-product and probably many more. – Gerry Myerson Jun 30 '22 at 04:04
  • More: https://math.stackexchange.com/questions/38671/two-sets-of-3-positive-integers-with-equal-sum-and-product and https://math.stackexchange.com/questions/1941758/sum-and-product-of-n-positive-integers and https://math.stackexchange.com/questions/1712821/triples-of-natural-numbers-with-same-sum-and-product – Gerry Myerson Jun 30 '22 at 04:10
  • ...and https://math.stackexchange.com/questions/2420033/n-numbers-with-their-sum-and-product-equal-to-each-other – Gerry Myerson Jun 30 '22 at 04:16
  • I saw most of those in my searching. Alas, all of them are limited to $\mathbb{Z}$, which sadly isn't that helpful. I seem to recall that one result in the integers was that you can start with any multiset of integers, then add a bunch of $1$s to get a solution for any integer other than a prime. – Eric Snyder Jun 30 '22 at 04:43
  • You should have mentioned your searches. – Gerry Myerson Jun 30 '22 at 12:52
  • @GerryMyerson That seems a bit odd to have as background/context? "Everything I found searching was limited to $\mathbb{Z}$, which makes them useless." The question specifically is working in a subset of $\mathbb{Q}$, so it seems silly to mention searches that turned up nothing related. I thought a base assumption was that the question writer had already searched and found little to nothing. (Assuming a well-written question.) – Eric Snyder Jun 30 '22 at 22:36
  • I have seen far too many counterexamples to that "base assumption". – Gerry Myerson Jun 30 '22 at 23:01
  • Sure, we've all had our share of "Welcome, your question makes no sense, here are the guidelines" comments. In this case, however, I worry that potential answerers/commenters just skimming will see eight links to other questions and assume this is solved... despite the fact that the links provide zero information germane to the question. – Eric Snyder Jul 01 '22 at 07:42
  • (Apologies if that sounded combative. Usually your comments and answers are very helpful.) – Eric Snyder Jul 02 '22 at 03:46
  • Note that $a,b,c$ are "price rationals" with $abc=a+b+c$ if and only if $A=100a$, $B=100b$, $C=100c$ are integers with $ABC=10000(A+B+C)$. I don't know whether this helps – I din't find anything useful by searching for sum divides product – but I do generally prefer working with integers to working with rationals. Oh, and if you want to be sure I see a comment intended for me, you should put @Gerry in it.) – Gerry Myerson Jul 03 '22 at 00:34
  • @GerryMyerson Yeah, that may be a path to examine, though the 10^4 (10000) is slightly intimidating! – Eric Snyder Jul 05 '22 at 10:45
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    I did find several questions here and on Art Of Problem Solving that looked at integer solutions of $ABC=(A+B+C)k$ for various values of $k$, but values of $k$ much smaller than, and with fewer divisors than, $10000$. Still, there may be something in those other questions that would be useful here. – Gerry Myerson Jul 05 '22 at 13:33

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This only answers your question about multiple solutions. There can be multiple solutions to a problem of this type (at least with three values):

$$(0.80)(2.50)(3.30)=.80+2.50+3.30=6.60$$ and $$(1.10)(1.50)(4.00)=1.10+1.50+4.00=6.60$$

[I found these using a computer search with Python.]

paw88789
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    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 $100$). – Eric Snyder Jun 29 '22 at 22:45