Math and Logic Puzzles

[Sudo_Nym]

I've got a Knights and Knaves, if it's okay for me to jump in.

[Sudo_Nym]

Do not post a puzzle if it is not your turn unless the person who won says it is OK. (I'll allow exceptions for if it takes an unreasonably long time to put up a puzzle, but "an unreasonably long time" has yet to be defined.)

Which is why I stopped to ask if it was okay if I jumped in. I'm not rude, you know.

[Sudo_Nym]

Anyway, the puzzle:

There are three gods, A, B, and C. In no particular order, one always tells the truth, one always lies, and one answers randomly. You are allowed to ask three yes-no questions of the gods (each question adressed to a specific god). However, while each god understands English, they will only answer with "Da' or "Ja"- one of which means yes, and one of which means no, but you don't know which is which. What three yes-no questions will allow you to identify which god is which?

[Sudo_Nym]

Ask one if the others are the God of Lies.
Then Ask the other guy if the others are the God Of Trust.
Then ask the Last Guy if the first guy is the God of Confusion.

Unfortunately, doesn't work.

[Sudo_Nym]

Also, just to clarify, you may ask a god more than one question, so long as you don't exceed three questions total.

[Sudo_Nym]

ask one god if he tells the truth and then ask him about each god in turn.

logik happens

Again, doesn't work. First of all, what if he's random? And second, his answers would be all "da" or "ja"

[Sudo_Nym]

Alright, I've been at this for almost an hour and I

think

my solution works. It kind of came from me eventually giving up on trying to figure out what 'ja' and 'da' actually are. After that, I figured I had to use them in my questions in some way that would get me an answer I could interpret without knowing their meaning. Here's my guess:

1. Ask God A, "Would you say 'ja' if I asked you if God B is the one that answers randomly?".

_________________________________________________

2a (if the answer to 1 is 'ja'). Ask God C, "Would you say 'ja' if I asked you if you are the one that answers only truthfully?".

3a (again, if 1 was 'ja'). Ask God C, "Would you say 'ja' if I asked you if God A is the one that answers randomly?".

_________________________________________________

2b (if 1 was 'da'). Ask God B, "Would you say 'ja' if I asked you if you are the one that answers only truthfully?".

3b (again, if 1 was 'da'). Ask God B, "Would you say 'ja' if I asked you if God A is the one that answers randomly?".

_________________________________________________

An answer of 'ja' to 2a or 2b means that the god asked is the truthful one. 'Da' to 2a or 2b means he is the liar. If 'ja' to 3a or 3b then A is the random one. If 'da' to 3a or 3b then the other one (not A and not the one you asked) is the random one. PoE tells you the third in any case.

Unfortunately, I don't think this works, but it's close. It fails because your first question isn't specific enough- putting aside the case where A turns out to be random, neither a Ja or Da answer will not be specific enough to be useful. You're correct, though, that in the correct series of questions, it doesn't matter what Da and Ja actually mean.

[Sudo_Nym]

All the cases are as follows:
If God A is truth, Ja means yes, and B is random, he'll answer Ja.
If God A is truth, Ja means no, and B is random, he'll answer Da.
If God A is lies, Ja means yes, and B is random, he'll answer Ja.
If God A is lies, Ja means no, and B is random, he'll answer Da.
If God A is truth, Ja means yes, and B isn't random, he'll answer Da.
If God A is truth, Ja means no, and B isn't random, he'll answer Ja.
If God A is lies, Ja means yes, and B isn't random, he'll answer Da.
If God A is lies, Ja means no, and B isn't random, he'll answer Ja.

In half the cases, a Ja answer means either A or B is random, and in the other half, it means either A or C is random. Thus, it doesn't actually provide you much useful information.

[Sudo_Nym]

Well, now I get to be embarrassed, I guess, since I misread your question and wound up with bad results. It looks like your question should work, then.

[Sudo_Nym]

Actually, after reading it correctly, Kage's solution actually is the intended one. I just misread it, and thought he was asking something different.

[Sudo_Nym]

I have another puzzle, I guess, going back to the Pirate game. To recap:

A group of pirates comes across a chest of 100 coins, and needs to divide it. But these pirates have an interesting system for dividing up treasure: each pirate has a number, and the one with the highest number picks a distribution. Then all the pirates vote on the distribution- on a tie or better, the distribution is accepted; but if the majority votes no, the proposing pirate is thrown overboard, and the next in line proposes a new distribution.
The pirates vote according to the following rules:
1) Pirates are not suicidal, and if voting one way would kill them, they always vote the other.
2) Pirates are greedy, and will vote in the manner that will maximize the number of coins they recieve, provided doing so does not violate rule 1.
3) Pirates are bloodthirsty, and will vote no, provided doing so does not violate rules 1 or 2.

As you saw before, the highest ranking pirate can actually snare a disproportionately large number of the coins. The question is, what is the largest number of pirates that can exist in the group, before the lead pirate can no longer survive?

[Sudo_Nym]

Is the answer 202?

Nope.

[Sudo_Nym]

Your logic works, but doesn't go far enough, Taz.

[Sudo_Nym]

No, mine is much higher.

[Sudo_Nym]

Yes, I am suggesting that. Remember, under my rules, the pirates will always vote yes, if voting no would guarantee that they die, even if they get no coins for it.

[Sudo_Nym]

Nope.

[Sudo_Nym]

Well, let others try first, I should think.

[Sudo_Nym]

I don't know. What's the maths?

[Sudo_Nym]

Using Tazaro's logic, you can show 200 pirates works by giving P200, P198, P196, ... , P2 1 coin each. Similarly, for 201 and 202 pirates, you would give a coin to every even-numbered pirate (sans P202 in the 202 pirates case, as he will vote yes on survival reasons). 203 pirates doesn't work, as he can buy the votes of 100 pirates and vote yes himself, but he needs 102 votes. 204 pirates, on the other hand, does work. He gives a coin to P202, P200, ... , P4, buying himself 50 votes. He votes yes himself, but so does P203. If P203 doesn't vote yes, then P204 is killed, making him the next to suggest a distribution, and as previously shown, there is no way for him to live in that scenario. This gives P204 102 votes, which is enough to survive. 205 doesn't work for the same reason as 203. 206 doesn't work as he can buy 100 votes, along with voting himself and getting a free vote from P205, but it's not enough. 207 gives you another free vote, but it also ups the vote amount required by 1. 208 pirates works, as you get one more free vote from P207. The number of pirates that allows the pirate with the highest number to survive are of the the form 200 + 2^x, where x is a non-negative integer. With 200 + 2^x pirates, (200 + 2^x)/2 pirates need to vote yes. (Technically, it's the ceiling of that, but I've proven the only case where 200 + 2^x would be odd.) To simplify, 100 + 2^(x-1) pirates need to vote yes. Buying 100 votes, along with getting 2^(x-1) - 1 free votes from one pirate below the top number down in addition to the highest-numbered pirate himself voting yes would give 100 + 2^(x-1) - 1 + 1 yes votes, or 100 + 2^(x-1) votes. As 200 + 2^x can be made infinitely high, the answer is infinite.

Yes. You can convince an infinite number of pirates to vote yes with only 100 coins, so long as there are exactly 200+2^n pirates in the group (201 and 202 being 200+2^0 and 200+2^1), by exploiting the fact that all pirates with numbers 200+2^(n-1)<x<200+2^n must vote yes in order to live, in accordance with rule 1.

[Sudo_Nym]

The question is, what is the largest number of pirates that can exist in the group, before the lead pirate can no longer survive?

The way the question is worded, the answer is 203. Because every number up to that, each pirate can make a reasonable deal to stay alive. Once there are 203 pirates in the group, the lead pirate can no longer survive. Even though the hypothetical Pirate 204 could survive, I think with the wording "before the lead pirate can no longer survive" the answer is actually 203.

Fair enough. I apologize if Taz feels cheated. But CO's answer is what's intended.

Proof by contradiction is a bit simpler:

If there were some number of pirates, X, such that proposals made by a pirate with a number higher than X will always be rejected, then if we have 2X pirates, then pirates X+1, ..., 2X should vote yes for whatever offer 2X makes since they will all die otherwise by assumption.

Remember rule three- pirate X+1 can't vote yes, if he has the chance to vote no and still live.

[Sudo_Nym]

You are the ruler of a medieval empire and you are about to have a celebration tomorrow. The celebration is the most important party you have ever hosted. You've got 1000 bottles of wine you were planning to open for the celebration, but you find out that one of them is poisoned.

The poison exhibits no symptoms until death. Death occurs within ten to twenty hours after consuming even the minutest amount of poison.

You have over a thousand slaves at your disposal and just under 24 hours to determine which single bottle is poisoned.

You have a handful of prisoners about to be executed, and it would mar your celebration to have anyone else killed.

What is the smallest number of prisoners you must have to drink from the bottles to be absolutely sure to find the poisoned bottle within 24 hours?

[Sudo_Nym]

1000?

That wouldn't be much of a math puzzle, would it?

[Sudo_Nym]

Given that it's a math and logic puzzle, I doubt you can count on getting lucky.

[Sudo_Nym]

Sudo, what is that orange thing in that person's hand in your avatar?

It's a cheeseburger.

999?

I get the feeling you aren't really applying math nor logic to this puzzle.

[Sudo_Nym]

Nope. You can test all the wine, and know for sure, with far less than 999.

[Sudo_Nym]

10 prisoners.

Divide the bottles into 10 groups of 100

Prisoner A drinks combined drops from all bottles in Group A
Prisoner B drinks combined drops from all bottles in Group B
etc.
Prisoner J drinks combined drops from all bottles in Group J

Lets say Prisoner J dies.

Prisoner A drinks combined drops from 12 bottles in Group J
Prisoners B-I each drink combined drops from 11 bottles in Group J

Lets say Prisoner I dies.

Prisoners A-C drink combined drops from 2 bottles in Group J-I
Prisoners D-H drink a drop from one bottle each in Group J-I

As stated, this would work, if you had all the time in the world. Unfortunately, it takes up to 20 hours for the poison to take effect, and you only have 24 hours to identify the poison.

You have to split it at the same time so that when, after 20 hours, people die, you can work out a common wine that each person had.

Sadly, I can't do the maths for that at the moment.

Yes, this is your hint. And there is a way to do it.

That overruns the three-guess limit.

Do people want me to give the answer, or leave it open for more guesses? I admit, I didn't really read the rules.

[Sudo_Nym]

You only need 10 slaves to test all the bottles. Assign each bottle a binary number, in order. Then make each slave take a sip from each bottle, according to the binary pattern- so if there's a 1 in the 1s place, Slave A drinks from it. A 1 in the 2s place means that Slave B drinks, and so on. With 10 slaves, you can generate 1024 different combinations, allowing a unique combination for each of the 1000 bottles. Then you wait to see what combination of slaves dies from the poison, and the bottle that corresponds to that binary combination must be the poision.

[Sudo_Nym]

A stark raving mad king tells his 100 wisest men he is about to line them up and that he will place either a red or blue hat on each of their heads. Once lined up, they must not communicate amongst themselves. Nor may they attempt to look behind them or remove their own hat.

The king tells the wise men that they will be able to see all the hats in front of them. They will not be able to see the color of their own hat or the hats behind them, although they will be able to hear the answers from all those behind them.

The king will then start with the wise man in the back and ask "what color is your hat?" The wise man will only be allowed to answer "red" or "blue," nothing more. If the answer is incorrect then the wise man will be silently killed. If the answer is correct then the wise man may live but must remain absolutely silent.

The king will then move on to the next wise man and repeat the question.

The king makes it clear that if anyone breaks the rules then all the wise men will die, then allows the wise men to consult before lining them up. The king listens in while the wise men consult each other to make sure they don't devise a plan to cheat. To communicate anything more than their guess of red or blue by coughing or shuffling would be breaking the rules.

What is the maximum number of men they can be guaranteed to save?

[Sudo_Nym]

Spoiler: My solution

The wise men devise a code before the king lines them up.

The first man will count up all the red hats in front of him. If it is odd, he will say "red", and if it is even he will say "blue". He has a 50/50 chance of survival.
The second man again counts all the red hats in front of him. If the first man said "red" and there is an odd number of red hats, his hat is blue. If the first man said "red" and there is an even number of red hats, his hat is red. If the first man said "blue" and there is an odd number of red hats, his hat is red. If the first man said "blue" and there is an even number of red hats, his hat is blue.

This continues down the line until the last 99 people are saved.

This is correct.

[Sudo_Nym]

Getting a running start, and deliver a devastating and visually impressive flying kick to the window, thereby shattering it?

[Sudo_Nym]

WHAT? That would totally get her out of the room.

[Sudo_Nym]

Plexiglass? You just don't watch enough Chuck Norris movies.

[Sudo_Nym]

1. Split the balls into groups of three, then weigh group 1 against group 2.

2a. If weighting 1 is different, then the different ball must be one of those eight- it's then possible that either the light group contains a ball that's too light, or the heavy group contains a ball that's too heavy. Place 2 balls from the light group and 1 ball from the heavy group on each side.

3a(i). If those weigh the same, then one of the last two must be too heavy. Weigh them against each other, and the heavy one is different.

3a(ii). If weighting 2a is different, then it either must be the possibly heavy ball on the lower side, or one of the possibly light balls on the lighter side. Weigh the two possibly light balls against each other. If they're the same, then the possibly heavy ball is the one; if they don't match, the lighter ball is the one you're looking for.

2b. If weighting 1 is the same, then group 1 and group 2 are all fine, and the ball must be in group 3. Weigh three known correct balls against three unknown balls.

3b(i). If weighting 2b is the same, then the ball must be one the last one. You can weigh it against any of the other balls, if you wish to determine if it's too light or too heavy.

3b(ii). If weighting 2b is different, then it must be one of those three. We already know if the ball is too light or too heavy, depending on how step 2b tipped, so weigh 2 of the 3 against each other. If they balance, then it's the last one. If they don't balance, then whichever corresponds to light or heavy, depending on step 2b must be the ball.

[Sudo_Nym]

There are five houses on a street, all in a row. Each house is a different color; furthermore, in each house lives a person of a different nationality, who has a different favorite drink, smokes a different brand of cigarette, and owns a different pet.

Hints:
1. The Brit lives in the red house.
2. The Swede has a dog.
3. The Dane drinks tea.
4. The green house is immediately left of the white house.
5. The owner of the green house drinks coffee.
6. The person who smokes Pall Malls owns birds.
7. The owner of the yellow house smokes Dunhills.
8. The man living in the center house drinks milk.
9. The Norwegian lives in one of the houses on the end.
10. The man who smokes Blend lives next to the man that owns cats.
11. The man who owns a horse lives next to the man that smokes Dunhills.
12. The man that smokes Blue Masters drinks beer.
13. The German smokes Prince.
14. The Norwegian lives next to the blue house.
15. The man that smokes Blend lives next to the man that drinks water.

So who on this block has a pet fish?

[Sudo_Nym]

Yeah, this was precomposed. I wanted a classic logic problem, but I don't know how to compose them.

[Sudo_Nym]

Spoiler: Answer

	House 1	House 2	House 3	House 4	House 5
Colour	Yellow	Blue	Red	Green	White
Nationality	Norwegian	Dane	Brit	German	Swede
Drinks	Water	Tea	Milk	Coffee	Beer
Smokes	Dunhill	Blends	Pall Mall	Prince	Blue Masters
Pet	Cats	Horse	Birds		Dogs

So the German owns the fish.

But I'll let SK have it


As for the other one... They all have an odd number of digits. Maybe each pair is a set of coordinates? I'll look into it.

Congrats.

[Sudo_Nym]

I have another puzzle. Just throwing that out there, if someone solves Quilford's and doesn't want it.

[Sudo_Nym]

If it hits a bird on the way, can I count the time it spends bouncing as part of its travel time?

[Sudo_Nym]

Damnit, I have a puzzle, and everybody keep claiming it before I have a chance.

[Sudo_Nym]

To the best of my memories, it's a touchdown as soon as he breaks the planes of the goal line. Assuming that when you said "tackled at the the 2", he wasn't downed there, then it's a touchdown. If tackled at the 2 did mean he was downed, then it's not a touchdown, because he was downed at the two.

[Sudo_Nym]

You've got 81 bags, and in each bag are 5 coins. 80 of the bags contain identical coins- all the same size and weight. However, one bag contains 5 coins which are slightly lighter than all the other coins. All 5 coins in this bag are too light by exactly the same amount.

You can take the coins out of the bags, and play around with them as much as you like. You've also got labels, and can label the coins and the bags how you like (for the purposes of the puzzle, you have infinite labels, and they're al weightless).

Using a standard two-pan balance, what's the fewest number of weightings you need to identify which bag is the light one?

[Sudo_Nym]

Luck doesn't really count, does it?

[Sudo_Nym]

Nope. 4 is the wrong answer anyway.

[Sudo_Nym]

It means the coins in one bag weigh less. The difference in weight is not enough to be dedicated just by feeling them, thus the scale. For the purposes of the riddle, simply looking at the coins isn't sufficient to be able to tell the difference.

[Sudo_Nym]

Actually, you only proved that you needed a maximum of four in the best case scenario, which is not how the game is played. Which is also why Junpei's answer of "one" didn't work. I'm looking for the system that will determine the light bag in the fewest number of weighings, in every possible scenario.

I talked to some friends at work about it, and I should clarify. When I say "standard two-pan balance", I mean a balance that's identical to the one I own, which tells you how much lighter one side is than the other. Which I mention, because it's important. So, free hint, I guess.

[Sudo_Nym]

I never said 1 was wrong. I said Junpei's logic was not good enough to identify the light bag with 100% certainty after just one weighing. If you can find a way to do it in just one, let me know.

[Sudo_Nym]

As you like, but 4 is still not correct.

[Sudo_Nym]

Remember, you can open the bags, and weigh the coins seperate from their bags. ANd remember that the light coins are all identical to each other- so they're lighter than the regular coins by exactly the same amount. That's important.

[Sudo_Nym]

but the answer is smaller than 4 then, because I proved 4 can find it with absolute certainty, correct?

If you like.

[Sudo_Nym]

Digi's answer is the correct one. By using the ratios like that, you can figure it out in exactly two weighings.

[Sudo_Nym]

Also, I'm sorry I left the part about knowing how much lighter the one bag is than the other. I assumed everyone had a balance like mine, I swear!

[Sudo_Nym]

Both men had insulted a Mafia Godfather in identical fashion, but since both men were identical, the sharpshooter thought the one man was actually both men.

[Sudo_Nym]

Hey, you asked for a reason why two identical guys could drink identical drinks and have only one turn up dead. I have done this, therefore, I am the winner.

[Sudo_Nym]

Obviously, both men were identical sharpshooters, both hired to kill themselves. But they're identical, so the one confused the other for himself.

[Sudo_Nym]

A paragraph puzzle! Game's had none of these yet. Bad grammer, huh? Hard to write! Can't express myself properly. Hard picking suitable words. Why? Is it just me? Identify rule to writing. Difficult problem: say why. Answering better means more win with you! Good luck, people!

[Sudo_Nym]

Alphabetical each sentence?

Words in each sentence are in alphabetical order?

Indeed, solution this.

[Sudo_Nym]

You're playing a new variant of pool- when you rack up the balls, you set a point value for each ball. The number of points each ball is worth when sunk is equal to its number on it, plus the numbers of every ball that was touching it when the balls were racked. As an additional rule, no two balls with consecutive values are allowed to touch. What's the maximum number of points you can score in this variant?

[Sudo_Nym]

536?

Edit: I assumed 15 balls are used.

15 balls are used, but you can get higher than this.

[Sudo_Nym]

627

.

There are three balls that will be counted thrice, nine that will be counted five times, and three that will be counted six times.

Put {1,2,3} as the ones that will be counted three times, and {11,13,15} as the ones that will be counted six times (they all touch, so they have to be two apart from each other).
It is then trivial to place the remaining balls in successful positions so the rules are satisfied and they are all counted five times.

Thus, (6*3) + (75*5) + (39*6), or 18+375+234, gives

627

.

Your logic is right, but your math is slightly wrong.

683, Sudo?

How do you figure? This is greater than what I have, so if you're logic is good, you may be one up on me.

[Sudo_Nym]

I'm going for 666.

This is correct. The center balls actually get counted 7 times, not 6, which was Digi's mistake.

[Sudo_Nym]

Since CES usually passes, I've got another.

For a huge Mafia Scum meet, I've set up a display of clocks, each one set to a different time, but not running, so the time displayed doesn't change. When I get back, though, I notice some of the clocks are missing, but I can't remember how many clocks I had to start with. The clocks currently on the shelf, from left to right, are set to 7:15, 7:30, 6:15, 11:35, 6:35, then a blank spot, 5:45, then a blank spot, and finally, 1:55.

How many clocks are missing, and what time should they be set to?

[Sudo_Nym]

That one went faster than I thought. Too many of you are familiar with semaphore.

[Sudo_Nym]

Here's a twofer:
A man walks 10 miles south, 10 miles east, and 10 miles north, and winds up back at his starting position. What color bear does he see?

A man flies a plane 2700 miles south, 2700 miles east, then 2700 miles north, and winds up back at his starting position. What continent is underneath him?

[Sudo_Nym]

The answer to the first is no bear at all- regardless of the fact that he'd have to be at the north pole, there are no bears living on the polar ice shelf, and you won't find any bears living within a thousand miles of the poll, let alone 10 miles. Besides that, polar fur isn't white, it's translucent. It only appears white as an optical illusion. So that part was answered correctly by theamatuer.

The second part, though, has only once correct answer, which hasn't been stated.

[Sudo_Nym]

The answer to the first is no bear at all- regardless of the fact that he'd have to be at the north pole, there are no bears living on the polar ice shelf, and you won't find any bears living within a thousand miles of the poll, let alone 10 miles. Besides that, polar fur isn't white, it's translucent. It only appears white as an optical illusion. So that part was answered correctly by theamatuer.

The second part, though, has only once correct answer, which hasn't been stated.

The same question has been asked in this thread three times already; there is no "one correct answer"... there are infinite possible locations.

Fine, be that way, spoilsport. Just trying to throw some counterfactuals out there, but you gotta be all "Infinite possible locations".

[Sudo_Nym]

Alright, here's more of a puzzle hunt one. I have a bunch of clocks, one on each shelf in a vertical line, set to the following times:

5:35
6:35
6:35
6:35
7:15
9:50
9:50
5:35
7:15
6:35
5:35
6:35
9:50
5:35
6:35
9:50
5:35
9:50
9:50
7:15

However, the clock on the bottom shelf is missing. Using your senses, can you figure out what the bottom clock should be set to?

[Sudo_Nym]

Fine, be that way, spoilsport. Just trying to throw some counterfactuals out there, but you gotta be all "Infinite possible locations".

I didn't mean to spoil your fun.

Nah, that's cool. Just giving you a hard time.

[Sudo_Nym]

The polar bear one was a no, because you won't find any bears. The south pole one was supposed to be South America, as the only landmass with any land at 2700 miles away, but that got mothballed.

[Sudo_Nym]

Alright, here's more of a puzzle hunt one. I have a bunch of clocks, one on each shelf in a vertical line, set to the following times:

5:35
6:35
6:35
6:35
7:15
9:50
9:50
5:35
7:15
6:35
5:35
6:35
9:50
5:35
6:35
9:50
5:35
9:50
9:50
7:15

However, the clock on the bottom shelf is missing. Using your senses, can you figure out what the bottom clock should be set to?

I worked hard to make this puzzle, guys.

[Sudo_Nym]

7:15?

Nope.

[Sudo_Nym]

This doesn't appear to be a semaphore problem, but rather a pattern problem. Unless I messed up, I can't make anything out of NAAAMOONMANAONAONOOM.

Might have to use all of your powers of observation...

[Sudo_Nym]

So you're saying we should smell out the problem?

Well, not quite...

[Sudo_Nym]

Taste?

No, but now you know which two are useless...

[Sudo_Nym]

Yes, metaphorically.

[Sudo_Nym]

I didn't mean to bring this to a dead stop...

Now that you've dotted right past the first layer of semaphore, I'm sure you'll dash right into the next layer, if only you keep your ear to the ground for clues.

[Sudo_Nym]

I arranged the clocks vertically for a reason, btw.

[Sudo_Nym]

This puzzle will be easier solved if I tell you that I'm using American Morse, not International; the key difference being that the code for O is . ., rather than - - -

[Sudo_Nym]

Well, he had most of the puzzle solved, I just want to help you move it along, ya know.

[Sudo_Nym]

It's not 5:35.

As another clue, it's not a pattern problem; trying to analyze the pattern won't get you anywhere. There is a logical reason, though.

[Sudo_Nym]

6:35 is correct. I must have stumped some people. Care to explain the logic before I do?

[Sudo_Nym]

Nice puzzling, Magua.

[Sudo_Nym]

Can't be, because it's not symmetric. If it's a cypher, it'd have to be one way. The problem is, if it is a cypher, we need to find a third string to be decoded.

[Sudo_Nym]

Or maybe where there are overlaps.

[Sudo_Nym]

You're the Captain on a submarine, when you recieve the following message:

TOP SECRET
Captain Mitchell
We have a spy in our midst and we believe he
is giving information to the Russians. We
have recently made a breakthrough and have
been able to narrow down our list of suspects
to one of the 252 members of your crew. We
have no exact information and no physical
description. We are convinced that you are
not the spy and as such we entrust you to
perform some covert investigation to aid us
in our attempt to catch the rogue operator.
We wish you good luck in this endeavour but
we must remind you to keep this secret. We
needn't remind you a refusal to follow these
instructions exactly to the letter will
result in an immediate courtmarshall.
Instructions for arrest:
1) If you fail to find the spy then, upon
docking you will ensure that none of your
crew can leave the ship.
2) If you do ascertain who the spy is then
you should arrest him, confine to the Brig
immediately & place him under 24 hour guard.
From this point on you should cease all
communications with us.

A minute later, you recieve this message:
Addendum to previous message:
D-C / T-H / A-A / O-H / C-A / X-K / W-A / O-F / X-D.

What should you do next?

[Sudo_Nym]

I admit that this is a puzzle I saw elsewhere on the internet. Don't spoil it!

[Sudo_Nym]

Congrats.

[Sudo_Nym]

It's your turn, if you want to go.

[Sudo_Nym]

Calculate the integral of 1/(cabin) d(cabin)

[Sudo_Nym]

P1 takes all 4 tens
Best hand P2 can make is 9 high straight flush; you respond by taking a royal flush
If P2 blocks all the royal flushs by taking higher cards, then you take a 10-high straight flush in any unblocked suit. Again, best hand P2 can take is 9-high straight flush, and you can make a 10-high.
Either way, P1 wins.

[Sudo_Nym]

Damn, ninja'd, but I was enthusiatic.

[Sudo_Nym]

Two women are sitting in a street cafe. The first woman says she has three daughters; the product of their ages is 36, and the sum of their ages is equal to number of houses on the other side of the street. The second woman says that's not enough information to figure out their ages. So the first woman says her oldest daughter has blue eyes. Now the second woman knows their ages.

How old are the three daughters?

[Sudo_Nym]

Spoiler:

I'm assuming that "the oldest daughter has blue eyes" gives away that there is an oldest daughter. since it isn't enough information, there have to be two ways to factor 36 so that the factors add up to the same thing... also, one of those ways has to have a tie for oldest. So the different ways to factor 36 with a repeated product:
1*1*36
2*2*9
3*3*4
6*6*1
and the second and fourth of these both add up to 13.

So they're 2, 2, and 9.

This doesn't preclude other solutions to the puzzle, but the fact that she knows their ages from that info implies that the solution is unique.

Yup. The version of this that I remember was the mathematician and the census taker, but same diff.

KageLord is indeed correct that implosion's answer is correct.

[Sudo_Nym]

Are the numbers important?

[Sudo_Nym]

Is the cabin actually a dwelling?

[Sudo_Nym]

Goddamn it, Digi.

[Sudo_Nym]

Is the guy with the gun a ghost and/or poltergeist?

[Sudo_Nym]

Henry fires a shot up into the air, and it gets lodged in the ceiling of his house. Years later, Lewis gets mad, punches a wall, dislodging the bullet, which then pierces his skull and kills him.

[Sudo_Nym]

Henry fires a shot up into the air, where it gets lodged into a tree branch. Years later, Lewis gets mad, punches the tree, dislodging the bullet, which then pierces his skull and kills him.

[Sudo_Nym]

Henry fires a shot up into the air, where it gets lodged into a bear. Years later, Lewis gets mad, punches the bear, dislodging the bullet, which then pierces his skull and kills him.

[Sudo_Nym]

Assume you have a chocolate bar which is m inches long and n inches wide, where m and n are natural numbers. You may break any existing piece into two equal halves, either vertically or horizontally. How many such breaks must you make to have only 1 square inch pieces?

[Sudo_Nym]

Correct.

[Sudo_Nym]

Who wants more classical logic problems?

Five people live on a street, and each one has a garden. Between the five, the gardens grow three different types of crop: fruits (apples, pears, nuts, and cherries), vegetables (carrot, parsley, gourds, and onions), and flowers (asters, roses, tulips, and lilies). Based on the following clues, what does each garden contain, where is it located, and who grows it?

1. All 12 types of crop are grown at least once.
2. Each garden has exactly four crops.
3. At least one garden contains a crop from each type.
4. Only one crop is grown in 4 different gardens.
5. Only one garden contains all three types of crop.
6. Only one garden contains all four crops of a single type.
7. Pears are only grown in the two gardens on the ends.
8. Paul's garden is in the middle, and does not contain lilies.
9. The person who grows asters doesn't grow any vegetables.
10. None of the gardens with roses also grow parsley.
11. The one who grows nuts also grows gourds and parsley.
12. The first garden on the street grows apples and cherries.
13. Only two gardens grow cherries.
14. Sam grows onions and cherries.
15. Luke grows exactly two types of fruit.
16. Exactly two gardens contain tulips.
17. Only one garden grows apples.
18. The only garden that grows parsley is next to Zack's.
19. Sam's garden is not on either end.
20. Hank's garden contains neither asters nor vegetables.
21. Paul grows exactly three types of vegetable.

[Sudo_Nym]

Fixes issued. 12 types of crop listed are correct, but I accidentally said there were five different types, rather than 3 different. I think it's otherwise correct, but I'm up late, so maybe some more mistakes. Be sure to let me know, and I'll repair.

[Sudo_Nym]

Who wants more classical logic problems?

Five people live on a street, and each one has a garden. Between the five, the gardens grow three different types of crop: fruits (apples, pears, nuts, and cherries), vegetables (carrot, parsley, gourds, and onions), and flowers (asters, roses, tulips, and lilies). Based on the following clues, what does each garden contain, where is it located, and who grows it?

1. All 12 types of crop are grown at least once.
2. Each garden has exactly four crops.
3. At least one garden contains a crop from each type.
4. Only one crop is grown in 4 different gardens.
5. Only one garden contains all three types of crop.
6. Only one garden contains all four crops of a single type.
7. Pears are only grown in the two gardens on the ends.
8. Paul's garden is in the middle, and does not contain lilies.
9. The person who grows asters doesn't grow any vegetables.
10. None of the gardens with roses also grow parsley.
11. The one who grows nuts also grows gourds and parsley.
12. The first garden on the street grows apples and cherries.
13. Only two gardens grow cherries.
14. Sam grows onions and cherries.
15. Luke grows exactly two types of fruit.
16. Exactly two gardens contain tulips.
17. Only one garden grows apples.
18. The only garden that grows parsley is next to Zack's.
19. Sam's garden is not on either end.
20. Hank's garden contains neither asters nor vegetables.
21. Paul grows exactly three types of vegetable.

1 (Hank) Pears, Apples, Cherries, Roses
2 (Sam) Onions, Cherries, Roses, Tulips
3 (Paul) Carrots, Gourds, Onions, Roses
4 (Zack) Tulips, Roses, Lilies, Asters
5 (Luke) Pears, Nuts, Gourds, Parsley

Not going to post a puzzle so keep solving the silly riddle or someone else post something new.

Correct

[Sudo_Nym]

Five children weight themselves in pairs. The totals for the various combinations, in some order, are 129 pounds, 125, 124, 123, 122, 121, 120, 118, 116, and 114. How much does each child weigh?

[Sudo_Nym]

Where's the flaw in this proof?

Claim: All horses are the same color.
Proof: Take a group of k horses. Obviously, the statement is true for k=1, since any horse is the same color as itself. Assume that any group of k horses is all the same horses. For any group of k+1 horses, remove any horse. Now you have a group of k horses, which are all the same color by inductive hypothesis. As this works for any horse you remove from the k+1 group, it must be the case that all horses in the k+1 group are the same color if horses in the k group are. Thus, since the statement is true for k=1, and k is true implies k+1 is true, it must be the case that all horses are the same color by induction.

[Sudo_Nym]

Well, you guys managed to home in on the 1 to 2 transition being the problem. Formally, as I understand it, the problem is one of transitivity. I'm assuming that in a k group, all horses are equivalent, so whichever horse I remove in the k+1 group is also equivalent to the other horses in the group. But in the k=2 group, removing any horse only proves that the remaining horse is equivalent to itself, which implies nothing at all about the other horse.

[Sudo_Nym]

Prove that a^p+b^p=(a+b)^p in any integral domain with prime characteristic p.

[Sudo_Nym]

If .9 bar isn't equal to one, then by density of rationals in the real line, there must exist a rational number x such that .999999...<x<1. Since no rational number such that this is true, this is a contradiction. Ergo .9999...=1.

[Sudo_Nym]

Robert's best chance is to murder the other twenty with a machete, I think.

[Sudo_Nym]

I posted this one a while back.

[Sudo_Nym]

With pidgeonholing, I can guarantee a subset that sums to 9 from a set of 10, but I don't get how you can guarantee 9 elements in the subset.

[Sudo_Nym]

If x+3 and x^2+3 are perfect cubes, then the product (x+3)(x^2+3)=x^3+3x^2+3x+9=(x+1)^3+8 is also a perfect cube. But 8=2^3, so we have (x+1)^3+2^3=k^3, which has no integer solution by Fermat's Last Theorem. Therefore, no such x exists.

[Sudo_Nym]

Alternately, (x+1)^3+8 is a perfect cube because it's the product of perfect cubes, and (x+1)^3 must be a perfect cube since x is an integer by assumption. The only way for this to be true is if (x+1)^3=-8 or 0, since all other perfect cubes are more than 8 apart. But then x=-3 or x=-1; if x=-3, then (-3)^2+3=12 is not a perfect cube, and if x=-1, then -1+3=2 is not a perfect cube. So, by contradiction, no such x exists.

[Sudo_Nym]

I know it's not my turn, but this came up in discussion at school, and I thought people here might enjoy it.

Two people are drawing marbles from an urn, which contains 50 black marbles and 50 white marbles. The first person draws a marble, and if it's black, he wins. Otherwise, he sets it aside, and the second player draws. Again, if the marble he draws is black, he wins. Otherwise, he puts it aside. The two players continue drawing marbles in this way until somebody draws a black marble and wins the game. What are the odds that the first person to draw will win?

[Sudo_Nym]

I know it's not my turn, but this came up in discussion at school, and I thought people here might enjoy it.

Two people are drawing marbles from an urn, which contains 50 black marbles and 50 white marbles. The first person draws a marble, and if it's black, he wins. Otherwise, he sets it aside, and the second player draws. Again, if the marble he draws is black, he wins. Otherwise, he puts it aside. The two players continue drawing marbles in this way until somebody draws a black marble and wins the game. What are the odds that the first person to draw will win?

Do we get a calculator? Do we get a fancy calculator which can do for loops?

Use whatever you want. You can also use Wolfram Alpha, if you want. Though that would seem to defeat the point.

[Sudo_Nym]

The probability that exactly one person gets the correct hat is 1/2013.
The probability that exactly 2013 people get the correct hat is 0.
The probability that at least k people get the correct hat is (2014-k)!/2014!

Spoiler: my reasoning

The probably that the first person got the correct hat is 1/2014; the probably that everybody else got the wrong hat given that first person got the correct hat is 2012/2013*2011/2012*...=2012!/2013!=1/2013. So, the probabilty that the first person got the correct hat and everybody else got the wrong hat is 1/2014*1/2013=1/4054182. This probably is for the first person, but it's the same probability for any member, since the order is arbitrary to the solution. So the probability that any one member got the correct hat and everybody else got the wrong hat is 2014/4054182=1/2013.

The probability that exactly 2013 members get the correct hat is 0- if 2013 people get the correct hat, then the only hat remaining for the last remaining person is the hat that was incorrect for everybody else; thus, it must belong to the last man. So if 2013 people get the correct hat, then 2014 people must have gotten the correct hat, so it's not possible for exactly 2013 people to get the right hat.

The probability that at least k members get the correct hat is 1/2014*1/2013*...*1/(2014-k+1)=(2014-k)!/2014! Note that, for consistency, this argument implies that the probability of everybody getting the correct hat is 1/2014!, and the probability that at least 2013 people get the correct hat is 1/2014!, so the probability of 2013 people getting the correct hat but not 2014 is 0.

[Sudo_Nym]

I think the problem is easier if you use polar coordinates. Define P=(0,0), and let x1=(Theta1, r1), and x2=(Theta2, r2).

If Theta1=Theta2+n(pi) for n=0, 1, 2..., then the shortest path is a line through the center, so the distance is r1+r2 if n is odd or abs(r1-r2) if n is even or 0.

Otherwise, the fastest way is to travel from the farthest point toward the center until you're on an arc that connects the two points. Let r=min{r1,r2}, and let Theta=min{abs(theta1-theta2), abs(theta1-theta2-2pi)}. Then, the d(x1,x2)=abs(r1-r2)+R*theta.

So we can generalize the above as d(x1,x2)=abs(r1-r2)+R*theta.

However, without loss of generality, assume r1>r2. Then, d(x1,x2)=r1-r2+r2*theta. If we were to travel from x1 to P, and then P to x2, we'd have gone a distance of r1+r2, so this formula is only good if d(x1,x2)<=r1+r2, or r1-r2+r2*theta<=r1+r2, or theta<=2.

So, d(x1,x2)=abs(r1-r2)+R*theta if Theta=min{abs(theta1-theta2), abs(theta1-theta2-2pi)}<2, and d(x1,x2)=r1+r2 otherwise.

[Sudo_Nym]

A chess king starts on any square of a chessboard. You must then make 64 moves, visiting each square on the board, and then returning to the starting square. Let a vertical or horizontal move be worth 1 point, and a diagonal move worth 0. What is the lowest number of points you can score on this loop?

[Sudo_Nym]

Proof?

[Sudo_Nym]

I think the question, as asked, is actually impossible:

Spoiler: reasoning

No stone in the group can have mass of an even number of grams, or the weight of the combination that contains just that stone is an even number, which is not allowed. So, every stone in the group must weight an odd number of grams. However, the sum of any two odd numbers is even, which is not allowed. Since each stone must weigh an integer number of grams, but can weigh neither an even number of grams or odd number of grams, such a set is impossible.

[Sudo_Nym]

I have 97:

97, 96, 95, 93, 90, 85, 77, 64, 43, 9

Although I inevitably overlooked something simple.

[Sudo_Nym]

I knew it'd be something simple.

[Sudo_Nym]

So the first in the sequence is x^.5, then (x^.5)^.5=x^.25, then x^.125, and so on (more generally, the exponent for the nth term is .5^n).

Case 1: x>=1
Then the sequence is monotone decreasing, and bounded below by 1, since k^.5<=k for all k=>1, and k^.5=k iff k=1. Thus, the sequence converges to 1 for x>=1 by monotone convergence theorem.

Case 2: 0<x<1.
Then the sequence is monotone increasing, and bounded above by 1, since k^.5>k for all 0<k<1. Thus, the sequence converges to 1 for 0<x<1 by monotone convergence theorem.

Thus, the sequence {x^(.5^n)} converges to 1 for x>0.

[Sudo_Nym]

What he's saying is that you can't move the limit inside the function- in this case, lim x^(1/2^n)=x^(1/lim 2^n), but it's not true generally, so you can't use it as a proof.

[Sudo_Nym]

The problem I'm having with the puzzle is the double counting- If there are 5 buckets, then two of them have to have even numbers of marbles if one of them does.

[Sudo_Nym]

If a tree falls in the forest, what color is it?

[Sudo_Nym]

You're all wrong. The tree is white, since you wind up back at the south pole.

[Sudo_Nym]

Here's one:
Suppose you've got a room full of strangers, and you ask them each their birthday. How many people need to be in the group before you have better than 50% odds that 2 of them share a birthday?

[Sudo_Nym]

Here's another prison one:

Suppose a prison has n prisoners on death row. Once per day, a random prisoner is picked (and the same prisoner can be picked more than once) and sent alone into a room, which is not visible from any of the prisoners cell, where there is a single light bulb, which randomly starts on or off. The prisoner may turn the light on or off or do nothing, as they wish. The prisoners may confer in advance, but may not communicate thereafter except through the lightswitch. At any time, a prisoner may assert that every prisoner has visited the room at least once. If he is right, all the prisoners are free; if he is wrong, all the prisoners are executed. What is the fastest way for the prisoners to win their freedom through this game?

[Sudo_Nym]

The most precision my algorithm can get is x = 0.002464167 radians, where there is a 1% difference between sin(x) and tan(x).

[Sudo_Nym]

I can't offer a proof, since I used MatLab, but 1?

[Sudo_Nym]

I probably did have an error, since I'm now getting -1. That said, I'm eager to see the proof.

[Sudo_Nym]

Here's a reasonably simple one.

What is the next number in this sequence: 1, 2/3, 1, 12/5, 8...

[Sudo_Nym]

Here's one I've been thinking about:

Say you've shuffled two complete decks of cards together, and draw the top five. What is the probability that you've drawn a straight flush?

[Sudo_Nym]

csc(2pi/3) should be 2/root(3). I'm assuming that the error is that the student mistook cosecant as 1/cos, rather than 1/sin.

[Sudo_Nym]

Take any n. Note that (1+1/n)^n = ((n+1)/n)^n, and that the n+1 element is (1+1/(n+1)^(n+1)=((n+2)/(n+1))^(n+1)=((n+2)/(n+1))((n+2)/(n+1))^n. Then, the ratio of the n+1 element to the n element is (n+2)/(n+1)*((n+2)/(n+1)*n/(n+1))^n = (n+2)/(n+1)*((n+2)/n)^n. Note that both elements are greater than 1 for any n, so the ratio is greater than 1 for every n. Thus, the sequence is non-decreasing.

[Sudo_Nym]

For any Eulerian circuit, all vertices must have an even number of edges, since you have to begin and end at the same vertex, or it's not a circuit. If any vertex had an odd number of edges, then there must be a second vertex with an odd number of edges (since any Eulerian path must have exactly 0 or exactly 2 vertices with an odd number of edges); then you would have to begin at one vertex and end at the other, which is not allowed, because then it is not a circuit. Thus, all vertices must have an even number of edges.

However, if any vertex has more than 3 edges, then it is impossible to visit every edge on a single visit to the vertex- you can start there and visit one edge "for free", then visit two more edges on the first visit to that vertex, but then you must visit the vertex a second time in order to visit the fourth edge. Thus, for the graph to be Hamiltonian, no vertex can have more than 3 edges.

Since the graph must now feature at least 1 vertex with more than 2 edges, but all vertices must have an even number of edges and no vertex can have more than 3 edges, we have a contradiction. Therefore, it is not possible for a graph to be non-trivial Eulerian and Hamiltonian.

[Sudo_Nym]

Let F(y) = √(x⁴+(y-y²)²). Then take F(y)/dy. Note that you can move the derivative inside the integral, since the function is continuous over [0, 1].

So, d/dy √(x⁴+(y-y²)²) = .5*(x⁴+(y-y²)²)^-.5*(2*(y-y²)*(1-2y)) = (x⁴+(y-y²)²)^-.5*(y-y²)*(1-2y) Note that dF/dy is discontinuous at 0, so we have an issue, but dF/dy = 0 for 1/2 and 1.

The integral evaluates to 0 if y = 0, and then the derivative is 0, so we have three critical points. So we can evaluate (and using a calculator for the crunching because it's late), we see the maximum is at y = 1.

[Sudo_Nym]

My original approach, which I could think of a good way to express so late at night, was like this: x is basically a function of y, and it's strictly increasing over [0,1], since a<b implies a^2<b^2 and then a^4<b^4 follows. y-y^2 is constantly increasing over [0,1), since y>y^2 on that interval, and then (y-y^2)^2 is also strictly increasing, since a<b implies a^2 < b^2. So the function must be strictly increasing over [0,1) so no y<1 can be the correct answer. Therefore, y=1 must be correct.

[Sudo_Nym]

That doesn't really work though.

Welcome to the world of ideal mathematics; where gamblers have infinite funds, cows are spherical, and every surface is frictionless!

[Sudo_Nym]

Welcome to the world of ideal mathematics; where gamblers have infinite funds, cows are spherical, and every surface is frictionless!

Is it OK with you if this goes in my sig?

Go for it.

[Sudo_Nym]

Easy. Sin(x) = sin(x)

[Sudo_Nym]

Sure. Sin(x) = rad(sin(deg(x))

[Sudo_Nym]

Here's one I saw, and it might have been this thread, so forgive me if it is, because I don't want to recheck 113 pages:

Suppose x is uniformly distributed on the interval [0, 1], y is uniformly distributed on [x, 1], and z is uniformly distributed on [y, 1]. What are the expected values of x, y, and z?

[Sudo_Nym]

Throw a smoke bomb, check both doors in the confusion.

[Sudo_Nym]

Shoot one of them in the foot with a gun. If he says "You shot me!", he's the truth-teller, and you can ask him which door to go through. If he says "You didn't shoot me!", then he's the liar, and you can ask the other one. If the stabby guy tries anything, shoot him too.

[Sudo_Nym]

An alternate problem for people who prefer statistics (and I don't want to forget it):

Suppose people's life expectancy is calculated like in EU4. That is, when you turn 40, you have a 2/3% chance to die immediately. For each passing year, your chance of death increases by 2/3% again (so you have a 4/3% chance to die at 41, a 2% chance to die at 42, and so on). What is then the expected length of your life under this model?

[Sudo_Nym]

Yes. You have a 4/3% chance of dying at 41, provided you didn't already die when you were 40.

[Sudo_Nym]

I don't mind use of a calculator, but I would like to at least see your reasoning, since my solution is close to yours but different.

[Sudo_Nym]

In any event, I did get 54 and change years as my answer.

[Sudo_Nym]

From what I can tell, Taz's answer is right. I'd like to offer an extension to the problem: suppose that you shuffle a pair of jokers into the deck, so the deck is now the original 16 cards, plus 2 jokers. What are the odds that it takes k draws to draw a duplicate card, if the two jokers are considered to be identical?

[Sudo_Nym]

The next three numbers are 1, 2, 3.

[Sudo_Nym]

Well, the key is the series is divergent, so there is no limit, and attempting to calculate one can therefore be problematic.

[Sudo_Nym]

Less new than you imagined, I think.

Anyway, the puzzle:

There are three gods, A, B, and C. In no particular order, one always tells the truth, one always lies, and one answers randomly. You are allowed to ask three yes-no questions of the gods (each question adressed to a specific god). However, while each god understands English, they will only answer with "Da' or "Ja"- one of which means yes, and one of which means no, but you don't know which is which. What three yes-no questions will allow you to identify which god is which?

[Sudo_Nym]

Didn't want to step on people's parade, since it had been a while, but I posted this puzzle and the preceding puzzle in thread a while ago. Figured it'd been long enough that new people were reading it, though. At least it confirms that the answer works out the same way.

[Sudo_Nym]

This is for the discrete case, but I think it works for the continuous as well:

So, first, let's assume a clock is basically a unit circle, that the hands are length 1 vectors, and designate 12:00:00 as zero- that is, at exactly 12:00:00, all three hands are pointing up on the y-axis, and this angle constitutes 0 degrees for all three.

Every time a second ticks, the second hand moves 1/60th of the way around the circle; that is 360 degrees/60 seconds = 6 degrees per second for the second hand.

The minute hand moves around the circle once per hour; that's (360 degrees/60 minutes)/(60 seconds/1 minute) = 1/10 degree per second for the minute hand.

The hour hand moves around the circle once per 12 hours; that's (360 degrees/12 hours)/(60 minutes/hour)/(60 seconds/1 minutes) = 1/120 degrees per second for the hour hand.

Also note that when calculating angles, the second hand and minute hand have to be calculated modulo 360.

Using a brute force algorithm, at 1:12:00, 4320 seconds have past since midnight. At that time, the second hand is at 4320 seconds * 6 degrees/second modulo 360 = 0 degrees; the minute hand is at 4320 seconds * 1/10 degree/second modulo 360 = 72 degrees; and the hour hand is at 4320 seconds * 1/120 degree/second = 36 degrees. Thus, the hour hand is perfectly bisecting the angle made by the minute and second hand at that time, so an example of this happening exists.

[Sudo_Nym]

I automatically excluded 12:00:00 and 6:00:00 as being trivial cases, and my algorithm automatically stopped when I found the first non-trivial solution. I could probably search for more.

[Sudo_Nym]

Disproof:

Let a = b = root(2). Then a and b are irrational, and a^b = root(2)^root(2). If a^b is rational, then the statement is disproved. If a^b is irrational, then let c = a^b, and then c^b = root(2)^root(2)^root(2) = root(2) ^ 2 = 2; thus c and b are both irrational, but c^b is rational. Either way, the statement is disproved.

[Sudo_Nym]

You're blindfolded at a table, with 20 coins; 10 are face up, 10 are face down, but you can't tell which are which, being blindfolded and all. Your goal is to divide the coins in two equal piles, each containing identical numbers of heads and tails. How can you do this?

[Sudo_Nym]

Indeed.

[Sudo_Nym]

OK. Here's a couple more, related to each other:

1. Consider an m×n matrix A. Prove that exactly one of the below tasks is possible:
   
   1. Finding the inverse of A.
      
   2. Finding a column vector in A that is a linear combination of the other column vectors in A.
   
   
2. Define A to be this matrix:
   

   Code: Select all

   ┌       ┐
   │ 1 2 3 │
   │       │
   │ 4 5 6 │
   │       │
   │ 7 8 9 │
   └       ┘

   
   Determine which of 1a or 1b is possible with this matrix and find what is requested of it.

1) Isn't this just as simple as pointing out that any square matrix with a determinant of zero is singular, and therefore not-invertable, and any matrix has a determinant of zero if and only if only of the columns is a linear combination of the others? Or are you looking for more?

2) Let A be the first column, B be the second column, and C be the third column. Then, -A+2B=C

[Sudo_Nym]

For any matrix, B is the inverse of A if A*B=I, where I is the identity matrix. det(I) is one, since it's the identity, so assume that det(A)=0. Then det(AB)=det(A)*det(B)=0, but det(AB)=det(I)=1. So a matrix with a determinant of 0 is not invertible, or we have a contradiction. Now we just have to prove that this non-invertible matrix has a column vector which is a linear combination of the other columns.

Assume that A has a column which is a linear combination of the other columns. Elementary row operations don't change the determinant of a matrix, so it's possible to reduce that column to a zero-column. Call that B. Then det(B) = 0 = det(A); B has a zero-column, so it's determinant is zero, and det(B)=det(A) since elementary row operations don't change the determinant. So if A is linearly dependent, det(A) = 0.

Assume that det(A) = 0. Then 0 is an eigenvalue for A, so Ax = 0 for some x != 0. That is, c(1)x+c(2)x+...+c(n)x = 0, and then -c(1)x = c(2)x+...+c(n)x where x!=0, so A is linearly dependent.

Thus, A is invertable if and only if det(A) != 0, and A is linearly dependent if and only if det(A) = 0.

[Sudo_Nym]

Say there's an island which has 26 villages, which are named A, B, and so on to Z. All these villages start off as pagan, but then 26 missionaries visit in turn. These missionaries, by coincidence, are also named A, B, and so on. Each missionary starts at the village with the same name as himself, converts the village to Christianity, then moves to the next village, until he gets to Z, and then he goes to A and continues around like this. However, any time the missionary attempts to convert a village that has already been converted, the villagers, feeling oppressed, kill the missionary and revert to paganism. That is, the missionary arrives in a village that has a name that matches his own, and travels the circle, converting any pagan village he enters, until he enters an already converted village, which becomes unconverted and kills the missionary. How many villages remain converted when missionary Z is killed?

[Sudo_Nym]

Spoiler: answer

0

Spoiler: response

Correct. Care to explain?

Do they arrive sequentially or at the same time?

It actually doesn't matter to the logic of the solution, so you can assume either.

[Sudo_Nym]

Plotinus's solution works for the assumption that they come at the same time. There is actually a general solution, where the missionaries arrive at unknown times and in unknown orders, but I wasn't sure how complex I should get.

[Sudo_Nym]

It sounds more like you're trying to make us do your homework.

[Sudo_Nym]

Challenge: arrange all the numbers 1-9 so that the first 2 digits are evenly divisible by 2, the first 3 digits are evenly divisible by 3, and so on. For example, 123456789 doesn't work, since 12 is divisible by 2, 123 is divisible by 3, but 1234 is not divisible by 4. There is a unique solution, though I don't require you to prove uniqueness, unless you really want to go the extra mile on that.

[Sudo_Nym]

381654729

is the correct answer. Solution was posted a lot faster than expected.

[Sudo_Nym]

Three logicians, oddly named A, B and C, are all wearing hats with a positive integer on them. None of them can see his own hat, but can see the other two, and are told that one of the integers is the sum of the other two. The conversation goes thusly:

A: I do not know what number is on my hat.
B: My hat says 15.

What numbers are on hat A and C?

[Sudo_Nym]

Yes.

[Sudo_Nym]

Felissan is correct.

[Sudo_Nym]

Prove that there exists two points on the equator that are on exact opposite sides of the globe but are the exact same temperature.

[Sudo_Nym]

Yep. And how's that for a real life application of the intermediate value theorem? I mean, I can't think of how this knowledge is useful in a practical sense, but now if you're at a party and there's no booze and somebody asks how math can apply to the real world, you have an example!

[Sudo_Nym]

I think it's something like the first three legs define a plane, so define a function of the distance between that plane and the fourth leg, and since the ground is continuous, there's a stable rotation by IVT. This proof obviously lacks rigor, but it's in there somewhere.

[Sudo_Nym]

Seven men are sitting around a table, drinking beer. Each person's mug contains some amount of beer (though the amount could be zero). As a weird ritual, the first person pours his beer into the other six mugs, dividing his beer exactly equally among them. Then each person around the table, in order, pours exactly one sixth of the beer in his mug into each of the other mugs. When the seventh man finishes pouring, the men discover that each mug now contains exactly the same amount of beer it started with. How much beer is in each mug, if the total amount of beer is 42 ounces?

[Sudo_Nym]

You shouldn't have to do a system of equations. If you logic it, there should be just the one equation you have to do.

[Sudo_Nym]

One paragraph.

[Sudo_Nym]

I believe that works.

Spoiler: My solution, if you're interested

First of all, note that because the end state is the same as the beginning state, we can imagine the scenario continuing indefinitely rather than ending after the 7th man pours without affecting the solution. So pick the man who has the least amount of beer in his mug before he starts pouring. Call him X, and call the amount of beer in his mug x ounces. Since X had the least amount of beer in his mug before pouring, each other person must have given him at least x/6 ounces. Because 6 people gave him at least x/6 ounces each, X must have recieved at least x ounces from other people. The only way he could have exactly x ounces, as he must by definition, is if his cup was empty and every other person gave him exactly x/6 ounces. That is, each cup, before pouring, must contain exactly x ounces, because otherwise there would be no way that each person could have given X exactly x/6 ounces. So, the cup that is about to pour must contain exactly x ounces, and the next cup to pour must have x ounces after recieving x/6 from the current person, so it must have x-x/6 = 5x/6 ounces, and so on down the line. So at any given time, each cup must contain x, 5x/6, 4x/6, 3x/6, 2x/6, x/6, and 0 ounces. And since they must sum to 42, x=12 as a simple summation. So the cups, from 1 to 7, must contain 12, 10, 8, 6, 4, 2, and 0 ounces.

[Sudo_Nym]

Spoiler: solution

The king must be on C3

As the board stands, either the king was on b3 or c2 with white to move, or it's black to move, since the black king is in check. The king can't have been on c2, since that puts it next to the opposing king. If the king was on b3, then it's in a double check, but the arrangement of the rook and bishop make that impossible, so the white king must have started black's turn in check from either the rook or bishop, which is impossible. So it must be black's move.

So let's go back a move. White must have just made a move that put black in check, because black couldn't have started white's turn in check. The bishop couldn't have moved, since the only places the bishop could have come from are on the same diagonal as the king itself, so the check must have been created by discovery, and the discovery was created by moving the king, since otherwise one of the white pieces would have had to remove itself from the board at the end of its move, which is impossible. The king must have moved from b3 to somewhere, creating the discovery. That somewhere must be to either a3 or c3, since any other move would leave the white king in check.

There must have been another piece on the board blocking one of the checks, then one of the two pieces moved to check the king on b3, and then the white king moved to either a3 or c3 to evade the check. It must have been a black piece that was captured by the white king's move, because otherwise a white piece disappeared. But then we have a problem, which is that the piece was blocking one of the two checks, but then has to wind up somehow on either a3 or c3 so that the white king could capture it, and there's no piece that can do it.

So we know that the king was on b3, black moved either the rook or the bishop to check the king, white made some move to evade the check without moving his king (since the king still has to be on b3 on the next move for the position to result) so the check was blocked by moving some other piece, black must have captured that piece somehow (since otherwise the piece vanished without being captured), and then the white king captured the piece with a move to either a3 or c3. There's no conventional piece that could have done that, since that would imply that the white piece blocked the check by moving to a3 or c3, which obviously doesn't block either the rook or the bishop. That implies that the black piece got to a3 or c3 by capturing a piece and simultaneously moving to an empty square, which can only be done by a pawn capturing en passant. So there must have been a black pawn on b4, which is the only square that blocks a check and can reach a3 or c3 by en passant capture, and then there must have been a white pawn on either a2 or c2 that enabled that capture. a2 is immediately ruled out, since en passant would require it to move to a4, which is occupied by the white bishop.

So, there must have been a black pawn on b4, and a white pawn on c2. The black pawn on b4 takes care of blocking the rook check, so black's original move must have been the bishop check. White blocked the check by moving the pawn from c2 to c4, black created the double check by capturing the white pawn en passant, then white captured the pawn on b3 with his king. So, the white king is on b3.

[Sudo_Nym]

Prove that the product of four consecutive integers can never be a perfect square.

[Sudo_Nym]

I did mean positive integers, and yes. I really should be more careful about that, since I think that's twice now I've failed to specify that I meant positive integers instead of just integers.

[Sudo_Nym]

Given that I have a bachelor's in Math, you'd think I'd be aware of these things. Alas :p Here's another:

Suppose you have n cubical building blocks, and you want to use these blocks to build the largest possible cube. After building, however, you find that you do no have a perfect cube- one of the sides has exactly one row too few. Prove that n is divisible by 6.

[Sudo_Nym]

Suppose there's a room that contains at least two martians, each martian has at least two fingers. You take a quick scan, and note that there are between 200 and 300 martian fingers in the room, and that each martian has the same number of fingers. From this knowledge, you then know exactly how many martians are in the room. So, how many martians are in the room, and how many fingers does each martian have?

[Sudo_Nym]

Suppose there's a room that contains at least two martians, each martian has at least one finger. You take a quick scan, and note that there are between 200 and 300 martian fingers in the room, and that each martian has the same number of fingers. From this knowledge, you then know exactly how many martians are in the room. So, how many martians are in the room, and how many fingers does each martian have?

Do you start with the knowledge of how many fingers each Martian has?

No. I realized I made a typo that made the problem unsolvable, so I went back and fixed it. Maybe better now.

[Sudo_Nym]

I suspect you want 17 martians by 17 fingers (only square of a prime number in that interval). However, I don't like the spuriousness of "you make a quick scan and get an absurdly large range of between 200 and 300 fingers total, but somehow you also know for a fact that every single martian has exactly the same number of fingers". Because if there are only a few martians, they must each have a fairly large number of fingers, so you'd have to count them individually to get that knowledge. And if there are many martians, how can you be sure you didn't miss one who played with fireworks as a kid, unless you go through all of them carefully?

Correct. And I accept no responsibility for the spuriousness of martian scanning technology.

Let A be the set of natural numbers from 1 to 15 inclusive. Let B and C be subsets of A, such that B has 13 elements, C has 2 elements, B union C = A and B intersect C is empty. Is it possible for the sum of the elements of B to equal to the product of the elements of C? (sorry if that's phrased inarticulately).

[Sudo_Nym]

That's something I like about mathematical puzzles like these; there isn't necessarily one "right" way to get the correct answer.

Spoiler: My Solution

Assume it is possible, then let x and y be the elements of C. Then 120-x-y=xy => 120 = xy+x+y = (x+1)(y+1) - 1 => 121 = (x+1)(y+1). The only factors of 121 are 1, 11, and 121, and 121 isn't in A, so x = 10 and y = 10, which is too many 10s.

Should I keep going with these, or let somebody else take a chance?

[Sudo_Nym]

Suppose you're in a maze that is an 8x8 grid, where you start in the bottom left square, and exit in the top right square. Each square on the grid has an arrow which point up, down, left, or right. On each square, you must move 1 square in the direction of the arrow, and then the arrow in the square you just left rotates 90 degrees clockwise. If this movement would bump you off the grid without reaching the proper exit, you instead stay where you are, the arrow in your square rotates, and then you move in that direction. Assuming that the arrows are originally distributed arbitrarily, are you guaranteed to eventually escape the maze, or not?

[Sudo_Nym]

@Sudo: Just a question for clarification: How should the case be treated that one reaches back at the entry point instead of the designated exit (which could happen after as little as one moves)?

Assume the entrance closes once the game starts, prior to the first move, and is thereafter treated as as a bounding wall.

[Sudo_Nym]

Nicely done, both of you.

[Sudo_Nym]

Suppose you flip n fair coins, and I flip n+1 fair coins. What is the probability that I flip more heads than you do?

[Sudo_Nym]

Quite right.

Two players are playing a game. In front of them are nine tiles, numbered 1 through 9, and take turns taking one tile apiece. The winner is the first person whose claimed tiles sum to 15. Does either player have a winning strategy?

[Sudo_Nym]

No, only hitting 15 is a victory. If neither player hits fifteen, by going too far over or under, the game is a draw.

[Sudo_Nym]

Spoiler: solution

Indeed, there is no winning strategy. If you arrange the numbers into a magic square, like:

8 1 6
3 5 7
4 9 2

Then it becomes clear that the game is functionally identical to Tic Tac Toe, which is always a draw with optimal play

[Sudo_Nym]

Spoiler: solution

Indeed, there is no winning strategy. If you arrange the numbers into a magic square, like:

8 1 6
3 5 7
4 9 2

Then it becomes clear that the game is functionally identical to Tic Tac Toe, which is always a draw with optimal play

The problem said "whose claimed tiles sum to 15", nothing about containing a subset which sums to 15. Tic Tac Toe allows you to have pieces on the board not in the winning 3. Thus, you can threaten multiple attacks at once, and you have to look more than 1 move ahead in order to force a draw. Furthermore, not all sums have only 3 pieces. For example: 1,2,3,9 would win but not on that magic square tic tac toe board.

So then it's actually a weaker version of tic tac toe. I never actually took game theory.

[Sudo_Nym]

Suppose I have one gallon of pure red paint, and one gallow of pure green paint, in containers large enough not to spill. I pour one pint of green paint into the red paint, pour two pints of the red mixture back into the green, pour half the green mixture into the red, then pour from whichever pot contains more into the other until both pot contain equal amounts of paint. Assuming I stirred thoroughly between each pouring, which of the two paints is more "pure"- the one that started green, or the one that started red?

[Sudo_Nym]

Right. Basically, the "trick" is that any red paint missing from the red pot must be in the green pot, and vice versa, so they have to be equally impure.

[Sudo_Nym]

You're sitting in your logic class, taking a multiple choice test, when you find that one of the questions has been rendered completely illegible by the copier, and the proctor doesn't have the key! The answers for the question, however, are as follows:

a) All of the below
b) None of the below
c) All of the above
d) One of the above
e) None of the above
f) None of the above

Which is the correct answer?

[Sudo_Nym]

I was going through some of my granddad's old papers, when I found an invoice that read:

72 turkeys: $x67.9y

Where the x and y represent numbers that are now illegible. What was the price of a single turkey?

[Sudo_Nym]

Let's try another game, and see if I'm capable of writing it down correctly. Suppose we've got a line of 50 coins, of various denominations. Each player takes turns claiming a coin from either end of the line, and the winner is the player who takes the highest value of coins overall. Does either player have a guaranteed winning strategy?