@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:
M3W3D13-ch-part-3-different-a-b-c.png
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. 🙂
