I've been trying to solve this problem for the last couple days but with no luck. I'm not even sure how I would translate the information into mathematical terms or equations. All help appreciated!
Here is the question translated from my native language:
"Dorotea, Fredrika, Nora, and Ulrika together have 117 glass marbles.The information about how many marbles each of them owns is extremely scarce. However, it is known that Dorotea has the most marbles and Ulrika has the smallest number. Fredrika has more marbles than Nora. Furthermore, it is known that one of the girls has 30% more marbles than one of the others, but it is unknown which two girls are involved. It is also known that one of the girls has 40% more marbles than one of the others, and that one of them has 50% more marbles than one of the others. How many marbles does each girl have?"
I've now found 4 whole numbers that satisfy the conditions, however my method felt clumsy and I would love to find a better solution. What I did now was writing up tables for the whole number solutions of f(x) =1.5x, g(x) = 1.4x and h(x) = 1.3x, and then finding similar numbers across the tables and combining them trying to find a sum of 117. If we call the girls D, F, N and U we get: D = 39, F = 30, N = 28 and U = 20. Here we have D having 30% more marbles than F, F having 50% more than U and N having 40% more than U.
There might be ways to speed it up, but your method is basically the way to do this. There are about a billion ways to divide $117$ into different piles. How do we reduce that number? Well, we want only 4 piles, and there are only $11,400$ ways to divide $117$ into 4 piles. So that's a big step.
Notice that one pile has to be $30\%$ more than another. So one person has $x$ marbles and another has $13x/10$, so $x$ must be a multiple of $10$. So two of the numbers have to come from the following set of pairs:
$$(10,13), (20,26), (30,39), (40,52), (50,65)$$
(where we stop because the next pair would make the sum too big.)
We can use the same trick with the $40\%$ fact. One person has $y$ marbles and the other has $14y/10$, so $y$ must be a multiple of $5$. So two of the numbers have to come from this set of pairs:
$$(5,7), (10,14), (15,21), (20,28), (25,35), (30,42),(35,49),(40,56),(45,63)$$
(and the next pair is too big.)
There are two possibilities. First, the solution might be a pair from the first list with a pair from the second list. It's not to hard to find $(30,39),(20,28)$.
The other possibility is that only three people are represented in the two lists. We can try to find a pair of pairs that share a number, and we get:
$$( (10,13),(10,14)), ((20,26),(20,28)), ((30,39),(30,42)), ((40,52),(40,56))$$
So the possibilities for $3$ of the numbers are
$$(10,13,14), (20,26,28), (30,39,42),(40,52,56)$$
The last triple is too large, and it's not hard to eliminate the other three.
EDIT: I see that if we assume that we're in the first case, then there are two people with $10x$ and $13x$ marbles and another two people with $5y$ and $7y$ marbles. So we get the Diophantine equation
$$23x+17y =117$$
which has only one positive solution $(3,4)$ which is the right answer. I don't see a way to eliminate the second case, however.