I don't understand... In the video, I thought the line was a number line?

[divinedolphin]

In the video, I thought the line was a number line? Meaning in this picture, a has a value of around 2, b has a value of 3, and c is around 5. But then later in the video, Prof. Loh says that if the lines for a and b are together, then b has a value of 0?

[debbie]

@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.