|patrickwonders||Mar. 11th, 2008 11:51 pm More Girl Scout Cookies!|
This is a follow-on question to my last post.
It's Girl Scout cookie season, still. The Samoas come twelve boxes to
a case. This diagram on the side of each case shows you how to
stack the cases on a pallet with fourteen per level:
Today's question: For what positive integers m and n are there
arrangements of an m x m grid of boxes and a single row
of n boxes oriented perpendicularly to those in the grid such that
they perfectly fill a square pallet with no gaps?
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|patrickwonders||Mar. 9th, 2008 11:32 pm Stacking Girl Scout Cookies|
Here's a simple algebra problem. I have a follow-on problem,
too. But, sticking with the name of the community, I'll save that
one for tomorrow.
It's Girl Scout cookie season (at least here in Minnesota). The
Samoas come twelve boxes to a case. This diagram on the side of
each case shows you how to stack the cases on a pallet with fourteen per
Today's question: Assuming the only gap is the one shown and all
cases are the same dimensions, can that arrangement fill a square pallet (without
hanging over the edges)?
And the answer is:
Let's say the boxes are w-units by h-units. Then, for the pallet to be square, it must be that: 3w + h = 3h. That means that w = 2h/3. So, then the question becomes, is there a value for h such that 5w = 10h/3 ≤ 3h = 9h/3.
As such, unless we can give the boxes a long side that's negative, we're out of luck on making the pallet square.
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|rdore||Mar. 3rd, 2008 06:31 pm|
One can easily embed the complex numbers into 2 by 2 real matrices by:
a + bi -> ( a b )
(-b a )
Given an n by n complex matrix A, one can replace each complex entry with a four real entries in this way. This gives a 2n by 2n real matrix B. (Both transformations are easily seen to be injective ring homomorphisms.)
Prove that det(B) = |det(A)|2.2 comments - Leave a comment
|joshua_green||Jan. 26th, 2008 06:52 pm That's an order!|
Determine all orderings on the field Q(x).7 comments - Leave a comment
|tweedcap||Jan. 7th, 2008 04:16 pm Yo Ho Ho!|
The pirates on a particular ship have an unusual way of dividing up the treasure after a successful plundering. First of all, there is a strict ranking of the pirates on the ship. There is a captain, a first mate, a second mate, and so on. The rank order of the pirates is well-known to everyone.
When there is treasure to be divided, the captain proposes a scheme for dividing it up (who gets what and how much). Then, all the pirates vote on whether this is fair. If more than half of the pirates approve the plan, that's what they do. Otherwise, the captain walks the plank and is never heard from again, every one else moves up the totem pole, and the new captain has to propose a scheme. This goes on until a vote passes (the captain does get a vote, but loses ties), or until there is only one pirate on the ship.
Assume that each pirate is motivated by survival instinct and by greed, so that each will act so as to get as much gold as possible, but it's better to get nothing than to walk the plank.
So suppose that you are the captain of a crew of 20 (including yourself), and you have just captured a chest of 1000 gold coins. What is your best move?
(Careful, guys and gals. This is a tricky question.)
I want to clarify what I meant by "motivated by survival instinct and greed". A pirate prefers any situation in which he survives to any situation in which he doesn't. Any scenario in which he gets 12 gold coins is preferable to any scenario in which he gets 11. And so forth. Given the choice between two scenarios which result in the same amount of gold, a pirate will choose the one that results in the most plank-walking. (Because then he's closer to the top; there's no danger of ending up with a crew that's too small, you can always get more pirates.) Furthermore, everybody understands everyone else's motivation.
This should be enough assumptions to handle the problem asked.
Okay, now that you've all had time to explore the possibilities (and there's lots of good analysis here already), let me answer some of the requests for more completeness. You're right, I haven't quite given you enough information yet. There are many ways to fine tune the details of this problem, and I wanted to give you all a chance to think about that for yourselves. (Feel free to explore the consequences of varying the assumptions, and let me know if you find something interesting!) My favorite two choices are the following:
Alternative 1) Given a collection of equally desirable divisions of the treasure, the captain will prefer the one which rewards more highly ranked pirates. (A crude way to think of this is the captain "buys votes", and he'll buy the cheapest votes first, and if two votes are equally expensive, he'll buy that of the more highly ranked pirate.)
Alternative 2) No pirate prefers any pirate to any other pirate, so faced with a choice between equally good distributions of the treasure, the captain chooses one at random (uniform distribution). Pirates love to gamble, so they'll reject a plan if their expected return (in the sense of probablility) from rejecting it is greater than or equal to the amount offered. For example, a pirate would NOT vote to accept 1 coin if the alternative is a 50-50 chance of getting 0 or 2 coins. (Either way the expectation is 1 coin, and all other things being equal, pirates prefer to make other pirates walk the plank.)
Current Location: Northside Public Library10 comments - Leave a comment
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|lyric_age||Mar. 5th, 2007 02:13 pm An interesting puzzle on bidding|
You are the bidder and you have to bid for a commodity, which has a certain worth to seller on a uniform distribution scale of 1 to 100. Of course, you don’t know what it is worth to seller and only the seller knows what its worth to him. If you bid more than what it is worth to seller, he will sell it to you, else not. If your bid is successful, after buying the commodity, you can sell it in market for 1.5 times what it was worth to seller.3 comments - Leave a comment
For eg, if it's worth 60 to seller, if you bid less than 60, the seller will not sell it to you. If you bid > 60 but less than 90 (1.5 x 60), you make a profit of 90 - your bid amount. If you bid > 90, you stand to lose bid amount - 90.
a) Now what is the optimal bidding price, if you are the buyer.
b) How high should be the markup (i.e selling price in market in terms of number of times the seller's worth) so as to make it optimal for buyer to bid?
|dkogan||Dec. 4th, 2007 08:52 pm puzzled gear|
Couple of weeks ago, I wanted to get a set of mugs with puzzles/riddles on them, to set out for guests. Looked around online, didn't find anything suitable, so figured I'd make something on a design-it-yourself site.Leave a comment
Then I thought others might be interested, so I made a whole bunch here.
Think this would interest people? Any ideas/suggestions? I could slap the same thing on shirts/bumper stickers/whatever, but not sure anyone would care.
(I also posted this to mental_workout, sorry if you're seeing it twice.)
|kentox||Oct. 14th, 2007 04:56 am BitTorrent Maths!|
This was an interesting problem which I played with a while ago, and it might be of interest to some of the rest of you. It concerns the BitTorrent protocol, which is used for the fast and reliable distribution of large files to lots of people. If you're unfamiliar with this particular protocol, I'll describe it after the problem statement.
The question is this:
Suppose I am the initial-seeder of a BitTorrent file, and it is a worst-case scenario: an infinite or near-infinite swarm that does not distribute the data particularly well. Such a swarm requests each of the N pieces (see below) at random, without regard to whatever has been previously uploaded: we'll assume that I don't discriminate between requests, either. How much data can I expect to upload before the swarm as a whole has one full copy of the file distributed among its members?I liked both my solution method and the answer, so I think others in this community might be interested.
If you've never encountered BitTorrent before, here's the general idea: the group of downloaders is called a "swarm," and they all contact some central information database, called a "tracker," in order to retrieve information about each other: they can then contact each other while they download the file. In particular, if one downloader grabs, say, the first 128 kilobytes of the file they're downloading, and some other downloader grabs the second 128 kilobytes, then they can download from each other, instead of the original source of the file. To facilitate this, the file is split into some number of pieces, and the person who's uploading the file provides metadata (hash codes) which allow your downloading client to verify the integrity of any given piece it has downloaded. This "trading" can remove a lot of stress from the original uploader, and since it removes the most critical speed-bottleneck, it makes downloads much faster.2 comments - Leave a comment
|rdore||Sep. 8th, 2007 02:51 am|
Given a sequence of complex numbers an converging to zero, must there be a sequence of 1s and 0s, in, so that (-1)i0 a0 + (-1)i1 a1 + (-1)i2 a2 + (-1)i3 a3 + ... converges?2 comments - Leave a comment
|joshua_green||Aug. 13th, 2007 10:28 am Analytic Goodness|
Can a nonconstant, entire function be real-valued in a neighborhood of some point?1 comment - Leave a comment
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