@divinedolphin Thanks for asking; this is a really neat solution method that reframes the question as something even more simple to calculate! This number line is a number line, but the locations of the $$a, b$$ and $$c$$ are shiftable, as if there are movable partitions placed at the $$\textcolor{blue}{\text{ blue }}$$ lines in the picture. So, the diagram that he drew might correspond to $$a = 2, b = 3, c = 4.5,$$ but there are many other possible values for $$a, b,$$ and $$c,$$ and this is exactly what we desire to solve for! The only requirements here are: You must keep the partitions in this order, with $$a$$ first, followed by $$b,$$ and followed by $$c.$$ (You cannot have $$b$$ first, followed by $$c$$ followed by $$a.$$ ) The partitions must lie at integer values on the number line ($$1, 2, 3, 4, 5,$$ or $$6.$$ ) Since $$0 \leq a \leq b \leq c \leq 6,$$ we can recast this in an amazingly simple way: The quantity $$a - 0$$ equals the number of coins allocated to the first pirate The quantity $$b-a$$ equals the number of coins allocated to the second pirate The quantity $$c-b$$ equals the number of coins allocated to the third pirate Here are some examples of different ways:   It is possible for $$a = 0,$$ meaning the first pirate gets none. It's also possible for $$b = a,$$ corresponding to the second pirate gets none. And it's possible for $$c = b,$$ corresponding to the third pirate gets none. If the question had stated instead that $$0 < a < b < c < 6,$$ then it wouldn't be as easy to turn this into a pirates-sharing-gold problem. It would still be doable, though! We would just need to make sure that each pirate get a gold coin to start with, and then apportion out the remaining coins. I hope this helped to make this more clear! Please ask if you have any more questions and I'll be more than happy to answer.