My alma mater has a math bowl every spring, and one of the questions when I was a student was how many gifts were given over the twelve days of Christmas. I got it wrong because the question WANTED the triangular answer and I gave them the tetrahedral one. Why must you bring this up again? ;)
(Though maybe I should send this to my advisor, who was the question writer/reader at said math bowl. Goodness knows he got enough stuff like this from me when I was a student...)
Here is my 12 days math post. It seems like this might be one of the few places that appreciates it.
Marvelous. You can even count -- with a small tad of multiplication -- the number of each type of gift. At first, I found the right angles of the rectangles of iso-colored balls a bit hard to observe. But when I imagined looking straight at the red, (partridge) edge -- that is, using a view point that's on the line connecting the tetrahedral centroid and the center of that edge, the tetrahedron being transparent -- then I see the rectangles.
In fact, you can look straight at the solid-colored rectangles from above either solid colored edge (the 12 partridges or the 12 drummers), and in switching between them, you observe the symmetry in the counts of the particular gifts: 5*8 gold rings; 8*5 milkmaids.
I'll take the golden rings, maids a milking, and ladies dancing, but for the rest I'll just take an equivalent amount of cash kthx.
In Rostov, Russia, 140 percent of listed voters reportedly took part in elections. Steven Moffat fans always vote twice. Nobody else votes twice. How many Moffat fans were there? :D
@Twicey: There are somewhere between 40% and 70% Moffat fans. 40 is everybody else votes once, and 70 if everybody else doesn't vote at all.
Neat — so one gift for each day of the year except Christmas itself (and February 29 when it exists).
Woah! Nice work Mike!
Erroneous, February 29th does not exist, has never existed, and will never exist. (exception: February 29th)
Is there a formula for tetrahedron numbers?