Let's start with an easy problem : What three positive integers have a sum equal to their product?
answer: (1,2,3), of course.
This puzzle can easily be transformed into a D4 problem:
For what values of k will the question "What k positive integers have a sum equal to their product?" have only one unique set of integers for an answer?
Clearly for k=2 the answer is unique: (2,2) and so it is for k=4: (1,1,2,4).
List all other values of k below 1000.
For starters, let's consider where only 2 of the numbers are not 1. List the values in descending sequence
First up, (m,2,1,...,1). The product = 2m, so the sum equals the product if there are 2m - m - 2 ones.
In which case there are m total terms. So one of these series works for all k = m, where m >= 2
Next up, (m,3,1,...,1). The product = 3m, so the sum equals the product if there are 3m - m - 3 ones.
In which case there are 2m - 1 total terms. So one of these series works for all k = 2m-1, where m >= 3.
In other words, k in (5,7,9,11 ...). So, there are no odd k that are unique greater than 3.
Next up, (m,4,1,...,1). The product = 4m, so the sum equals the product if there are 4m - m - 4 ones.
In which case there are 3m - 2 total terms. So one of these series works for all k = 3m-2 where m >= 4.
In other words, k in (10,13,16,19, ...).
In fact, just by considering (m, n, 1,....1) we can rule out any k = (n-1)m - (n-2) where m >= n
When n = 5, we can rule out (17,21,25,29...) These are all odd and have already been ruled out already
there is no need to consider odd n
When n = 6, we can rule out (26, 31, 36, 41, 46, ...)
When n = 8, we can rule out (50, 57, 64, 71, etc.)
Do far, we are down to 2,3,4,6,8,12,14,18,20,24,20,32,38,42,44,48,54 (up through 60)
I guess that we need to consider more terms than 2 that are not equal to 1
A start
