I Love Math Multi-Cache

The Overview:
So you have the final coordinates to the cache already. It’s a 50-caliber ammo can. How hard can it be, right? Well… there is a “combination lock” on the cache container. Locks seem to have this funny way of making things harder than they ought to be, don’t they?

First of all, so-called “locker combinations” are ubiquitous in high school culture. However, it is a misnomer. They are, in fact, locker permutations. See, the order of those three numbers matters. Below are the means to calculate the three numbers of the locker combination for the lock that is attached to the final cache container. (That’s right folks, you’ll have to determine the order once you get there in order to open up the lock.)

Also, you have been coordinates given for three other containers. Each of the three containers will have helpful mathematical tools or notes that will greater facilitate your in-the-field calculations. (This is why it has been listed as a multi-cache and not a mystery cache). If you find that doing the mathematics at home is proving to be quite difficult, consider looking for the help containers. I recommend bringing some scrap paper and a writing tool, but the use of a calculator will not be necessary – it is the conceptual set-up that is difficult, not the calculations themselves. Each of the questions were specifically designed to look quite hard at first, but to allow for the discovery of a simple and elegant solution.

Now onto the calculations. Make sure that you proceed carefully and cautiously. I wouldn’t want any miscalculations to set you off course.

The Easy Hard Calculation: N 42^o 41.208 W 073^o 45.903
One of the numbers in the combination is based on two probabilistic properties of sets of elements. The number of arrangements of a set equals the number of ways for which each of the elements within that set can be ordered. For example: { A, B, C, D, E, F } can be reordered as { B, C, F, E, D, A }, as { A, B, C, F, E, D }, and so on. The number of derangements of a set equals the number of arrangements for which no element remains in its original position. Using the example above, { B, C, F, E, D, A } is a derangement whereas { A, B, C, F, E, D } is not. For a set of four elements, calculate the number of arrangments that are not derangements.

The Medium Hard Calculation: N 42^o 41.299 W 073^o 45.926
Finally, what’s so special about the number 1 x 344,827,586,206,896,551,724,137,931? What about 2 x 344,827,586,206,896,551,724,137,931? What about 3 x 344,827,586,206,896,551,724,137,931? Well, there's a number that can multiply 344,827,586,206,896,551,724,137,931 that is quite interesting and you’ll know it when you see it; and when you see it, that multiplier will be another one of the lock’s numbers.

The Very Hard Calculation: N 42^o 41.308 W 073^o 45.983
Finally, the last problem is based on a vector calculus property that I will help you through (see notes below). There is a three-dimensional shape called a parallelopiped. There are 3 pairs of parallelogram faces that are respectively both congruent and parallel. Basically, it’s a rectangular prism wireframe without necessarily having right angles. The coordinates of the eight corners of our parallelopiped are: { (0,0,0), (1,3,1), (2,8,5), (5,5,2), (3,11,6), (6,8,3), (7,13,7), (8,16,8) }. Determine the volume of the parallelopiped to get a number for the lock.

Notes: there is a vector calculus property that shows that one may pick any corner of a parallelopiped and measure from the reference frame of that corner the 3 respective vectors needed to get from there to the adjacent corners. (For example: from (3,11,6), one of the vectors is found using collated subtraction: (3,11,6)—(2,8,5)=(1,3,1), which just also happens to be a different corner of the figure.) Then, those three vectors can be viewed together as a matrix whose determinant is a measure of the volume of the parallelopiped generated by those 3 vectors.

For example:

To calculate the determinant of a 3x3 matrix, you may have to research a little bit. However, I can assure you that the help container includes a method much easier than what the first 10 hits on google.com just turned up.

Final Advice:
You can use resources such as mathworld to guide you, but don’t overwork yourself. It is perhaps worth it to just show up and try your best. See how far you can get on the first try. Remember, there is helpful information to be found at each of the locations for this cache. Most of all, I hope that it will be a learning adventure for you. Good luck.