Playing with markers (bored during finals)

[BMAD]

While my students are taking their final exam, i found myself playing with my whiteboard markers. I have 23 markers in all. One of my many habits is that I enjoy snapping my markers together to make long line segments. Unfortunately 3 of my markers will not snap together properly meaning that at best I have three line segments. therefore each line segment is considered a set marker length out of the total markers. E.g. a marker group of 17 markers has length 17/23.

Now the question: If i create three marker lengths using all of the markers, and multiply the length of one line segment by the reciprocal of another by the average marker length of all three

meaning for example:

1st marker group: 17/23

2nd marker group: 2/23

3rd marker group: 4/23

average marker group: 23/69

17/23 * 23/2 * 23/69 = 391/138, approx 2.83

what combination would give me the smallest value? largest value? Is there a process or means to generalize this for n markers?

[Pickett]

So, isn't the average always going to be 1/NUMBER_OF_GROUPS? In this case, the "average marker group" is always going to 1/3 (aka 23/69).

As a result, that part of the equation can effectively be ignored...since it's constant (we'll just factor it in at the end to get the final values)

This leaves us with ONE_GROUP * INVERSE_OF_ANOTHER_GROUP...

In order to find the maximum, you would always want the INVERSE to the when the group was of length 1...because that means you would always have NUM_MARKERS/1 as one of the multipliers. Which means you just want to choose the largest group as your first one.

Therefore, to find the MAXIMUM value of n markers, given m groups, it would be

GROUP 1: (n - m + 1) / n

GROUP 2: 1 / n

...

GROUP m: 1 / n

Your maximum value would be (n - m + 1) / m...which in your case (n=23, m=3), would be 7.

To find the minimum, you would just do the exact opposite (choose the largest group as your reciprocal, and a group of ONE as your base...which is the same grouping as the maximum...which means your minimum value would be 1 / (m * (n-m+1))...which in your case above would simply be 1/63 = 0.015873...

So basically, you can always use the grouping I have listed above to get both the maximum and minimum values...