12 Days Of A Math Riddle

I hate math problems almost as much as I hate the 12 Days of Christmas. I hate these things for different reasons. With math, it’s because I’m bad at it. I survived high school algebra and precalc by the skin of my teeth and for college, I was able to take a philosophy logic course for math credit (and even then, I still got a D).

I generally don’t like Christmas music anyway, because it seems to constitute a genre of its own and yet it’s entirely stagnant, repetitive and a blend of nostalgia and tradition that seems to exist solely to perpetuate itself. The 12 Days of Christmas is the most repetitive song of them all, which is why it’s my least favorite.

So why am I mentioning these two things?

I’m not sure where I first heard this problem (probably some long-forgotten math class), but here goes: assume that the song lyrics are literal and that you actually receive a partridge in a pear tree on the first day and the second day and the third day and so on. Assume that each gift is considered a singular item (a piper piping is a single gift, even though you’re getting both a pipe and a guy to play that pipe for you). Which of the 12 different gifts will you have the most of at the end of the twelve days?

The answer is posted after the break:

Partridges: 1 × 12 = 12

Doves: 2 × 11 = 22

Hens 3 × 10 = 30

Calling birds: 4 × 9 = 36

Golden rings: 5 × 8 = 40

Geese: 6 × 7 = 42

Swans: 7 × 6 = 42

Maids: 8 × 5 = 40

Ladies: 9 × 4 = 36

Lords: 10 × 3 = 30

Pipers: 11 × 2 = 22

Drummers: 12 × 1 = 12

Total = 364

You’ll notice that the numbers form a parabola, peaking at the middle before descending.

I’m not much of a math guy but I thought that was pretty interesting and worth sharing. Incidentally, it’s also a good way to explain why it’s a good idea to invest money in a retirement fund when you’re younger, rather than waiting and trying to make the savings up later in life. Also, it’ll give you something to ask your relatives during Christmas dinner after someone says something awkwardly political.

You’re welcome.

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2 thoughts on “12 Days Of A Math Riddle”

  1. It’s easy to imagine this using a rectangle with a set perimeter (in this case 26), being adjusted by changing the amount on each side. So a 12×1 square, then an 11×2 square. With this in mind you can simply use two equations.

    A=L*W
    L+W=13
    A=L*(13-L)=13L-L^2

    We can therefore maximize by taking the derivative: dA/dL, and setting it to 0

    dA/dL=13-2L=0
    L=6.5

    Not surprising, as the general rule is the greatest area for a given perimeter is that of a square. (Actually, this is the way I solved it in my head, but wouldn’t really be considered rigorous from a mathematical standpoint).

    On a side note, this is technically not a “Bell Curve,” but a parabola.

    Sorry if I made you hate math even more 😛

    1. Damn! I couldn’t remember if it was a bell curve or a parabola, so I took a guess. I guessed wrong. Edited the post.

      And no, this comment didn’t make me hate math more than I already do. Actually, I was expecting a math related post would create an impressive comment from the the PhD aspirant I know reads my blog. I was not disappointed in that regard.

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