bonanova

Ed is dead. You question four suspects, of which one is guilty. Each suspect makes exactly one false statement and exactly one true statement. Alex: I did not kill Ed. Either Bob is the killer or none of us is.Bob: I did not kill Ed. Duke is the killer.Chuck: I did not kill Ed. Bob is lying when he says Duke is the killerDuke: I did not kill Ed. If Alex did not kill him, then Bob did.Whodunnit?

Fifty school children lined up at random in five rows and ten columns. No two of the children have the same birthday. Of the oldest children in each column Pete is the youngest. Of the youngest children in each row Jane in the oldest. What is the probability that Pete is older than Jane?

1. Flamebirde 2. BMAD 3. phil1882 4. bonanova 5. 6. 7. Backups 8. 9.

I'll give you four numbers and you give me two dozen. You can use + - * and / and ( and ) as often as you like. I was going to give you 2, 2, 8, 8, but it occurs to me that (2 + 2) * 8 - 8 = 24 is not challenging. So instead I'll give you 3, 3, 8, 8. Have fun!

Never played Mafia. Would be willing, if we do a basic game first, to learn.

I have the general idea. All the numbers, though ... ? Thinking.

Just occurred to me. Did you permit zeros?

I think there is only one answer. Here are the details of what I did. See if I missed something. In any event, I'm marking the puzzle solved. Good work.

No.

Nice work. But since Rainman knows the product, he knows at least one of these is not the right number.

The solution contains the assumption: I'm assuming that during the redistribution phase, I will know whether my opponent chose jar 1, 2, 3 or 4. I am trying to reconcile that with the red portion of the OP I was with you as to there being no way to gain an advantage until BMAD added this to the terms of the problem: After the first person chooses, you have the option to redistribute the marbles again. Given, that you get to redistribute the marbles, it is reasonable to assume you can tell which jar was chosen and then act accordingly. With this order of events: I fill the jars the way I want them Because the jars are not transparent my opponent has no clue about their contents My opponent picks one I know which one he picked, either because I was watching or because I now have access to all the remaining jars and marbles. I re-fill the jars the way I want them. I am blindfolded The jars are shuffled I pick one, while blindfolded there is a strategy.

That is an equivalent question. Nice. Along with the requirement that the center of the square can be arbitrarily positioned.

Good point. (no pun intended.) That would be a degenerate case - a square with side zero. And it wouldn't help to draw BD first. You'd get AC' collinear with C, with the same result.

I totally agree that stools with legs like that would be problematical. Happily very few stools are made that way. The set-up part of the question is intended to call to mind the generally well known property of three-legged stools (long ago milking stools were made with three legs for this very reason) that they do not rock, even if placed on uneven (e.g. barn) floors. The question is the sentence with the question mark at the end of it. To elaborate: Can a four-legged stool (also) be positioned (arbitrarily) over a mildly uneven surface in such a way that it is steady - i.e. will not rock? Equivalently, can a 4-legged card table (with nominally equal legs) be made to rest steady, without rocking, when positioned arbitrarily over a mildly uneven floor? The question is not meant to be tricky or out of the box, only to inspire some analytical thinking. Feel free to reply if something is still not clear.

Plasmid, does this cover the problem cases you noted?

No math required.

Your proof did not go "wrong": 1.902 IS "greater than or equal to" sqrt(3). What others have said is that 1.902 cannot be "equal to" sqrt(3). As several people now have shown, the best lower bound for M/m is 1.902, where equality can be achieved.

The solution contains the assumption: I'm assuming that during the redistribution phase, I will know whether my opponent chose jar 1, 2, 3 or 4. I am trying to reconcile that with the red portion of the OP

What does single exclusion mean?

At the aforementioned k-man took only imperceptible offense at Y-san's mainly-innocent comment. But he did retire brieflly to the library to concoct some good-natured revenge on the lady. Returning moments later to the main room he found her in conversation with Rainman, and he put to them a puzzle of his own: I've put a number into a sealed envelope here, he said, and it looks like this. And he showed them the sequence ABCDCBA. But of course my copy has numbers for the letters, he explained. I'm going to tell Y-san the sum of the seven digits, and I'll tell Rainman their product. Bonanova is not the only one who can resurrect an old puzzle form - you may already have seen puzzles of this type. It will be helpful if you have. You can discuss what you know between yourselves, but be careful not to disclose too much. First to come up with my seven-digit number gets a coveted bonanova gold star. Then k-man added, here's a clue: 9 >= A >= B >= C >= 0. and 9 >= D >= 0. Y-san spoke first, to Rainman: I have only a faint clue about what your number is. I have an even worse idea of what yours is, Rainman replied, this is difficult. I know, Y-san mused. At this point, your number is still uncertain. Rainman replied, now I know k-man's number! What is it?

Is it correct to conclude The hands begin rotating CW, all pointing at 12, at midnight Each hand reverses direction each time it approaches 12, from either direction The clock may have been running for an arbitrarily long time when the photograph was taken

Truth be told, the puzzle is a flawed attempt to present the "non-race" version, which I posted as separate problem once answers were posted here. I mean all the values from 1-9. My bad.

Well, you named this "paint color." Paint is formed of pigment, where CMYK is relevant. In the RGB space, I didn't find any of them remarkable.