bonanova

Ten minutes for the bunch ... Quickie 1 Quickie 2 Quickie 3

Having second thoughts about your answer, or difficulty seeing it.

The number abc has a decimal representation of ax102 + bx101 + cx100. But let's change things around a bit, and drop the base ten. We could then write abc = a2 + b1 + c0 and ask what values of a b c satisfy the equation. There actually may not be a solution. But if we play with the exponents a bit, we might come up with some numbers that do work. Try these (The numbers are in order, smallest to largest.) ab = a2 + b3 cd = c2 + d3 efg = e1 + f2 + g3 hij = h1 + i2 + j3 klm = k1 + l2 + m3 nop = n1 + o2 + p3 qrst = q1 + r2 + s3 + t4 uvwx = u1 + v2 + w3 + x4 yz@$ = y1 + z2 + @3 + $4 There is a shorthand notation. These can also be written ab (2,3), cd (2,3), efg (1,3,5), ... yz@$ (1,2,3,4). So here's a bonus challenge: abcdefgcc (4, 3, 8, 5, 7, 9, 0, 8, 8)

What fraction of triangles are obtuse? Provide a proof.

In case it helps, here's a clarification and restatement of the question. There are moments when it is not possible to identify which hand is the minute hand. An "ambiguous moment" is when that fact makes the time uncertain. How many ambiguous moments occur in any 24-hour period? (Assume we always know the angles of the hands as accurately as needed.)

I have seen neat puzzles where the answer is flat out obvious but (1) wrong or (2) has a simple proof that hides in the shadows. This puzzle, and the one about tiling a hexagon with diamonds, are type (2).

Closer but not all have been found.

@Molly Mae I gave your description a point and will mark it as solved in a couple days to allow some time to offer a proof.

The minute hand of a clock was lost and had to be replaced. Unfortunately only some extra hour hands were available for use. At how many moments in a day did the repaired clock, with identical hour and minute hands, give an ambiguous time? (Assume AM and PM could be deduced by other means.) Explanation: At noon (and midnight) the time was certain, even though the hands coincided. Also, a few minutes after these times, in fact at most other times, it can be deduced which is the minute hand. But there are certain hand positions where that's not possible. How many ambiguous moments occur in the time interval [Midnight, Next Midnight)?

@CaptainEd I agree with your reasoning but my numbers different enough for me to recheck them. @plasmid The fact you're that close to the Captain, I suspect my numbers are the incorrect ones. What I have convinced myself of, though, is that

@Molly Mae Consider that the sheriffs are simply talking to each other, using statements like, "my pair of suspects has this property: (describes property.)" Would generating and using common key agreements be doable with such low-complexity statements?

Good start, but there are more.

True, and that is a proof by construction, getting you a point. There is an unexpected proof that comes from looking at the hexagon from a slightly different angle, literally. Can you find it for the win?

Can you prove that's minimal? Can you think of a necessary property for your red squares?

For each n it's a fresh start. So really it's just one question: For any (every) n, can n red dots and n blue dots be joined pairwise with n line segments no two of which intersect?

You have three cards in front of you labeled in some order 1, 2 and 3, and three card-sized bins labeled with a letter that signifies Left, Middle and Right, in the order shown. The cards, all of which remain face up, have been placed at random in one or more bins, in such a way that, if any cards are stacked, only the top card is visible. And, if only two cards are visible, one cannot tell which bin holds the hidden card. +-----+ +-----+ +-----+ | | | | | | +---| L |---| M |---| R |---+ | | | | | | | | | +-----+ +-----+ +-----+ | +---------------------------------+ You are to devise an algorithm that ends with all cards in the L bin with 1 on top and 3 on the bottom, in a bounded number of moves. Each move in the algorithm consists of taking a single card from the top of one pile and placing it on the top of another (possibly empty) pile. An independent observer will keep track of things and will signal completion. Your algorithm, on the other hand, may not keep track of things. Each move it makes must take its cues only from the visible information at each point in the process. Specifically, a move cannot depend on any invisible cards, its previous move or the number of moves it has made. Further, it must be able to begin from a random starting point. A move might take the form "If a 2 is visible, move it one bin to the right." In that case, the 2 card would move to L if it happened to start in R. In that sense, the configuration functons as ring, with "move to the right" meaning CW, and "move to the left" meaning CCW. To restate: For any configuration with one, two or three visible cards, distributed in any order in L, M and R, your algorithm must specify the next move to make. And since your algorithm can't distinguish an L stack of 1 2 3 from 1 3 2, it will be told when it has (successfully) completed.

There was a killing last night at the *gasp* Double-R-Bar Ranch out between Dry Gulch and Tombstone. Each town's sheriff has investigated a list that contains the eight men who are the only possible suspects. And kudos to them both, because already each sheriff has a short list of just two suspects. They intend to compare their lists via telephone call and, if they agree on exactly one person, an arrest will be made. Unfortunately some of the town-folk know the names of the original eight suspects, and they instead intend to lynch the culprit, if they can identify him. Wait, there's more. They have tapped the sheriffs' telephone. If, by listening to the sheriffs' conversation, the mob can identify the culprit with certainty, the bad guy will be lynched before he can be arrested. How can the sheriffs, who have never met*, discuss their findings, opaquely to the eavesdropping lynchers but with clarity to each other, so that the mob is left in the dark, while both sheriffs end up knowing the culprit -- and thus allow either sheriff to make the arrest? *Specifically, they have no commonly-held information to use as an encryption key. (If something is not clear, please ask.)

Hi WorldDev and welcome to the Den.

I'm missing I think I understand your analysis, but wonder how you would implement this. Given they sit in a circle and guess (or pass) privately,

You are given a huge white nxn checkerboard and tasked with painting it red, one square at a time. You find the task tedious, so you quit after painting m of the squares red, and then you give the job to your assistant Bob. Bob is not a workaholic either, so "compassionately" you tell him he only needs to paint a square when he finds that it has two sides that border with red squares, regardless of who painted them. That is, Bob cannot quit painting until no white squares having two red borders remain. Clearly by wisely choosing your squares, and if m is large enough, you can compel Bob to paint all of the remaining squares. Clarifications: To border another square an entire side must be shared, not just a corner. If two red squares touch diagonally, Bob must paint the white square(s) that they both share a side with. If m = 1, or if you've painted squares sufficiently distant from the others, Bob won't have to paint any squares. Prove there is a least amount of work you must do (smallest m) that makes Bob complete the job.

A regular hexagon is divided into 2n equilateral triangles. Pairing triangles that share an edge produces diamond shapes with three distinct orientations, as shown. Prove that any n-diamond tiling of the hexagon will use the three types in equal numbers.