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# It's Math-Puzzle Time

Registered User regular
edited April 2008
It's that time of year again (Ok, so I made that part up) which means that it's time to play the "let's post amusing mathematical puzzles, oddities or anomalies game". Why? Because we can, and who doesn't love maths?

And to begin, here is the first mathematical puzzle of the thread.

---

There are three train stations, A, B and C, B sitting equidistant between A and C. Next to stations A and C there are Klimpy's chain restaurants. Given that stations A and C are in commercial districts of the city and B sits in the only nearby residential district the restaurants at A and C are almost entirely reliant upon business from people who have caught the train from B.

Customers, knowing that the distance between their station and the stations either side do not plan which Klimpy's they will eat at, instead they arrive at the train station and simply catch the next train, it's it's going to station A, they eat at that restaurant, if it goes to B, they eat at that restaurant.

However, Klimpy's central management discovers that the Klimpy's restaurant at station A does about three times as much business as the restaurant at station C.

Assume that the same number of trains travel to each station each day, that the gap between any particular two trains traveling to the same station is 20 minutes and that the times at which the customers show up to the train station is random/evenly distributed throughout the day.

Explain.

Apothe0sis on
Provide sample data to the Traitor project here || What is Traitor?
SODOMISE INTOLERANCE
Tide goes in. Tide goes out.
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## Posts

• Registered User regular
edited April 2008
Doc wrote: »
The solution to this is available on the 'net, so no cheating!

A prison warden decides to play a game with the inmates. If they can win, they all go free. If they lose, they all get stuck in there for life. The prisoners can all talk to each other to come up with a strategy before the game starts, but not after. The game is as follows: a prisoner will be randomly selected daily. He will enter a room where he sees two switches. He must toggle exactly one of the two switches and then exit the room. To "win," a prisoner must go to the warden and say "All prisoners have visited the room." If he is correct, they win. If not, they lose.

Again, after the game starts, the prisoners cannot communicate with each other. What's a good strategy for them to come up with before the game starts?

Can prisoners be selected more than once?

Lose: to suffer defeat, to misplace (Ex: "I hope I don't lose the match." "Did you lose your phone again?")
Loose: about to slip, to release (Ex: "That knot is loose." "Loose arrows.")
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• Registered User, ClubPA regular
edited April 2008
Linden wrote: »
Are the switches known to be stable across days? Do they start in known positions? Can the number of days passed be counted by the prisoners?

Let's say they start both down/off. The only people that mess with the switches are the prisoners, and the prisoners are free to count days passed. But a prisoner can be selected multiple times, possibly even twice in a row (as it's random every day).

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• Registered User
edited April 2008
Doc wrote: »
The solution to this is available on the 'net, so no cheating!

A prison warden decides to play a game with the inmates. If they can win, they all go free. If they lose, they all get stuck in there for life. The prisoners can all talk to each other to come up with a strategy before the game starts, but not after. The game is as follows: a prisoner will be randomly selected daily. He will enter a room where he sees two switches. He must toggle exactly one of the two switches and then exit the room. To "win," a prisoner must go to the warden and say "All prisoners have visited the room." If he is correct, they win. If not, they lose.

Again, after the game starts, the prisoners cannot communicate with each other. What's a good strategy for them to come up with before the game starts?

I've seen this puzzle before. For it to be solvable, the prisoners must be given enough information beforehand to be able to describe the different possible positions of the switches. That is, they must be able to make a list of instructions for themselves that refers to switches as either in position A or position B. On or Off, Left or Right, Up or Down - it doesn't matter what the two states are, but the prisoners must be able to walk into the room and know whether each switch is in position A or B. Example: what if the switches are not labeled, and do not turn anything "on" or "off" that the prisoners can see? What the first prisoner walks into the room on day one only to discover that switch 1 flips Left/Right while switch 2 flips Up/Down? What if the switches are buttons that get pressed and change colors from Red to Green with each push? etc.

This is a great puzzle though. Number of prisoners doesn't matter. Solution is easier if no prisoner can be called more than once, but it can be solved even if any prisoner can be called any random number of times. It's also easier if you know the initial state of the switches (both down/off, as Doc posted here) but it can even be done without knowing the initial state. All in all, rockin hard puzzle!

I'll leave the solution to somebody else and post one of my favorite tricky puzzles.

[SIGPIC][/SIGPIC]
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• Registered User, ClubPA regular
edited April 2008
Absurdist wrote: »
Doc wrote: »
The solution to this is available on the 'net, so no cheating!

A prison warden decides to play a game with the inmates. If they can win, they all go free. If they lose, they all get stuck in there for life. The prisoners can all talk to each other to come up with a strategy before the game starts, but not after. The game is as follows: a prisoner will be randomly selected daily. He will enter a room where he sees two switches. He must toggle exactly one of the two switches and then exit the room. To "win," a prisoner must go to the warden and say "All prisoners have visited the room." If he is correct, they win. If not, they lose.

Again, after the game starts, the prisoners cannot communicate with each other. What's a good strategy for them to come up with before the game starts?

I've seen this puzzle before. For it to be solvable, the prisoners must be given enough information beforehand to be able to describe the different possible positions of the switches. That is, they must be able to make a list of instructions for themselves that refers to switches as either in position A or position B. On or Off, Left or Right, Up or Down - it doesn't matter what the two states are, but the prisoners must be able to walk into the room and know whether each switch is in position A or B. Example: what if the switches are not labeled, and do not turn anything "on" or "off" that the prisoners can see? What the first prisoner walks into the room on day one only to discover that switch 1 flips Left/Right while switch 2 flips Up/Down? What if the switches are buttons that get pressed and change colors from Red to Green with each push? etc.

I'll leave the solution to somebody else and post one of my favorite tricky puzzles.

Right, they are physical switches with an up and a down. This can be known ahead of time.

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• Registered User regular
edited April 2008
Doc wrote: »
The solution to this is available on the 'net, so no cheating!

A prison warden decides to play a game with the inmates. If they can win, they all go free. If they lose, they all get stuck in there for life. The prisoners can all talk to each other to come up with a strategy before the game starts, but not after. The game is as follows: a prisoner will be randomly selected daily. He will enter a room where he sees two switches. He must toggle exactly one of the two switches and then exit the room. To "win," a prisoner must go to the warden and say "All prisoners have visited the room." If he is correct, they win. If not, they lose.

Again, after the game starts, the prisoners cannot communicate with each other. What's a good strategy for them to come up with before the game starts?

I think I have a solution for this, although it may not be the fastest one.
Spoiler:

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• Registered User regular
edited April 2008
That won't work if someone is called more than once.

My SteamID Gamertag and PSN: TheLawinator
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• Registered User regular
edited April 2008
Yes it will, because
Spoiler:

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• Registered User regular
edited April 2008
That won't work if someone is called more than once.

No, it's fine because if they have already switched the important switch they switch the other one and the control dude ignores it. Of course, the expected length of time for this whole method to work is some enormous times, surely longer than the shortest sentence any of the prisoners is serving, but I think it would work.

Lose: to suffer defeat, to misplace (Ex: "I hope I don't lose the match." "Did you lose your phone again?")
Loose: about to slip, to release (Ex: "That knot is loose." "Loose arrows.")
·
• Registered User regular
edited April 2008
That won't work if someone is called more than once.
Spoiler:

and her knees up on the glove compartment
took out her barrettes and her hair spilled out like rootbeer
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• Registered User
edited April 2008
The solution to like every math puzzle ever is on the internet at this point, so no cheating if you want a good challenge! Some people get this right away, but it was really hard for me. Like, keep-me-up-all-night-and-then-eureka-while-showering-the-next-morning kind of hard. PM me for a hint.

What is the next number in this sequence?

1, 4, 7, 12, 15, 18, 21, 24, 27, ?

[SIGPIC][/SIGPIC]
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• Registered User regular
edited April 2008
Actually, I think that will work, but it will take a really long time.

But I don't mind, as long as there's a bed beneath the stars that shine,
I'll be fine, just give me a minute, a man's got a limit, I can't get a life if my heart's not in it.
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• Registered User regular
edited April 2008
Corlis wrote: »
Actually, I think that will work, but it will take a really long time.

That's why I said it's probably not the fastest, although that's partly due to the relatively large example value of X. If we were talking about 10 prisoners, my method would probably take only a few months.

And if the prisoners don't have to flip any switches, there's an easy way to cut down the time:
Spoiler:

EDIT: Actually, that doesn't work because it would require flipping two switches on the same visit. Never mind.

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• Registered User
edited April 2008
Absurdist wrote: »
The solution to like every math puzzle ever is on the internet at this point, so no cheating if you want a good challenge! Some people get this right away, but it was really hard for me. Like, keep-me-up-all-night-and-then-eureka-while-showering-the-next-morning kind of hard. PM me for a hint.

What is the next number in this sequence?

1, 4, 7, 12, 15, 18, 21, 24, 27, ?

I love ambiguous sequences!
Spoiler:

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• Registered User
edited April 2008
cyphr wrote: »
Absurdist wrote: »
The solution to like every math puzzle ever is on the internet at this point, so no cheating if you want a good challenge! Some people get this right away, but it was really hard for me. Like, keep-me-up-all-night-and-then-eureka-while-showering-the-next-morning kind of hard. PM me for a hint.

What is the next number in this sequence?

1, 4, 7, 12, 15, 18, 21, 24, 27, ?

I love ambiguous sequences!
Spoiler:

Jeez, some people.

Ok, to clarify, what is the next number in this sequence that does not require the use of multiple rules or "if...then" statements. Since, you know, you can obviously define any sequence via an infinite number of functions if you allow the use of compound functions.

1, 4, 7, 12, 15, 18, 21, 24, 27, ?

[SIGPIC][/SIGPIC]
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• Registered User
edited April 2008
Alright alright, point taken. But now I'm also going to be up for a while thinking about it. Bastard. :-p

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• Registered User
edited April 2008
Here is another tricky one!

What's the next number in this sequence?

10, 0, 3, 20, 16, 51, 32, 67, 74, ?

[SIGPIC][/SIGPIC]
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• Registered User
edited April 2008
I'll post my own while I ponder yours.

1, 11, 21, 1211, 111221, 312211, 13112221, ?

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• Registered User
edited April 2008
Cantide wrote: »
Doc wrote: »
The solution to this is available on the 'net, so no cheating!

A prison warden decides to play a game with the inmates. If they can win, they all go free. If they lose, they all get stuck in there for life. The prisoners can all talk to each other to come up with a strategy before the game starts, but not after. The game is as follows: a prisoner will be randomly selected daily. He will enter a room where he sees two switches. He must toggle exactly one of the two switches and then exit the room. To "win," a prisoner must go to the warden and say "All prisoners have visited the room." If he is correct, they win. If not, they lose.

Again, after the game starts, the prisoners cannot communicate with each other. What's a good strategy for them to come up with before the game starts?

I think I have a solution for this, although it may not be the fastest one.
Spoiler:

Technically that's not a solution, as technically the problem isn't solvableâ€” every prisoner is not necessarily going to be called into the room after any amount of time.

That is a "good strategy", though

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• Registered User
edited April 2008
cyphr wrote: »
I'll post my own while I ponder yours.

1, 11, 21, 1211, 111221, 312211, 13112221, ?

A classic! This sequence was the first of it's type that I had ever come across, and I never did get it. I had to ask for the answer. Since then, of course, I look for solutions of a similar type in any stumper puzzle.
Spoiler:

[SIGPIC][/SIGPIC]
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• Registered User
edited April 2008
Technically that's not a solution, as technically the problem isn't solvableâ€” every prisoner is not necessarily going to be called into the room after any amount of time.

That is a "good strategy", though

He's got you there, Doc. You need to stipulate that the prisoners and the universe they live in are immortal.

hehe

[SIGPIC][/SIGPIC]
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• Registered User, ClubPA regular
edited April 2008
Cantide wrote: »
Doc wrote: »
The solution to this is available on the 'net, so no cheating!

A prison warden decides to play a game with the inmates. If they can win, they all go free. If they lose, they all get stuck in there for life. The prisoners can all talk to each other to come up with a strategy before the game starts, but not after. The game is as follows: a prisoner will be randomly selected daily. He will enter a room where he sees two switches. He must toggle exactly one of the two switches and then exit the room. To "win," a prisoner must go to the warden and say "All prisoners have visited the room." If he is correct, they win. If not, they lose.

Again, after the game starts, the prisoners cannot communicate with each other. What's a good strategy for them to come up with before the game starts?

I think I have a solution for this, although it may not be the fastest one.
Spoiler:

That is a solution. It does take a really long time.

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• Registered User
edited April 2008
Doc wrote: »
Cantide wrote: »
Doc wrote: »
The solution to this is available on the 'net, so no cheating!

A prison warden decides to play a game with the inmates. If they can win, they all go free. If they lose, they all get stuck in there for life. The prisoners can all talk to each other to come up with a strategy before the game starts, but not after. The game is as follows: a prisoner will be randomly selected daily. He will enter a room where he sees two switches. He must toggle exactly one of the two switches and then exit the room. To "win," a prisoner must go to the warden and say "All prisoners have visited the room." If he is correct, they win. If not, they lose.

Again, after the game starts, the prisoners cannot communicate with each other. What's a good strategy for them to come up with before the game starts?

I think I have a solution for this, although it may not be the fastest one.
Spoiler:

That is a solution. It does take a really long time.
Does that mean there's a better solution? I must know.

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• Registered User regular
edited April 2008
Cantide wrote: »
Doc wrote: »
The solution to this is available on the 'net, so no cheating!

A prison warden decides to play a game with the inmates. If they can win, they all go free. If they lose, they all get stuck in there for life. The prisoners can all talk to each other to come up with a strategy before the game starts, but not after. The game is as follows: a prisoner will be randomly selected daily. He will enter a room where he sees two switches. He must toggle exactly one of the two switches and then exit the room. To "win," a prisoner must go to the warden and say "All prisoners have visited the room." If he is correct, they win. If not, they lose.

Again, after the game starts, the prisoners cannot communicate with each other. What's a good strategy for them to come up with before the game starts?

I think I have a solution for this, although it may not be the fastest one.
Spoiler:

Technically that's not a solution, as technically the problem isn't solvableâ€” every prisoner is not necessarily going to be called into the room after any amount of time.

That is a "good strategy", though

If every prisoner isn't called in at some point, it's unwinnable anyway.

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• Registered User
edited April 2008
The game as described is not necessarily winnable, yes.

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• Registered User regular
edited April 2008
Doc wrote: »
Cantide wrote: »
Doc wrote: »
The solution to this is available on the 'net, so no cheating!

A prison warden decides to play a game with the inmates. If they can win, they all go free. If they lose, they all get stuck in there for life. The prisoners can all talk to each other to come up with a strategy before the game starts, but not after. The game is as follows: a prisoner will be randomly selected daily. He will enter a room where he sees two switches. He must toggle exactly one of the two switches and then exit the room. To "win," a prisoner must go to the warden and say "All prisoners have visited the room." If he is correct, they win. If not, they lose.

Again, after the game starts, the prisoners cannot communicate with each other. What's a good strategy for them to come up with before the game starts?

I think I have a solution for this, although it may not be the fastest one.
Spoiler:

That is a solution. It does take a really long time.
Does that mean there's a better solution? I must know.

I've come up with a solution, but it seems a little slow (but not as slow as the previous one). This assumes that they HAVE to flick a switch every day.
Spoiler:

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• Registered User regular
edited April 2008
The traditional prisoner puzzle only has one switch, and a lot of work has been done on it:

http://www.segerman.org/prisoners.pdf
http://www.ocf.berkeley.edu/~wwu/papers/100prisonersLightBulb.pdf

Edit: Obviously, these links can be considered massive spoilers for the prisoner problem. You've been warned.

Most methods can be extended to two lightbulbs (switches) in some respect; the first paper has a small section on two switches, as well as a bunch of other variants.

Interestingly, to my knowledge, no-one has been able to prove what the best solution is.

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• Registered User
edited April 2008
Nobody can solve my puzzlers. Do I win something?

10, 0, 3, 20, 16, 51, 32, 67, 74, ?

1, 4, 7, 12, 15, 18, 21, 24, 27, ?

[SIGPIC][/SIGPIC]
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• Registered User
edited April 2008
Spoiler:

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• Registered User regular
edited April 2008
This is most assuredly not my series, and math geeks will probably see why this is funny.

2, 5, 877, 27644437, 35742549198872617291353508656626642567, 359334085968622831041960188598043661065388726959079837 ... ?

Provide sample data to the Traitor project here || What is Traitor?
SODOMISE INTOLERANCE
Tide goes in. Tide goes out.
·
• Registered User
edited April 2008
Apothe0sis wrote: »
This is most assuredly not my series, and math geeks will probably see why this is funny.

2, 5, 877, 27644437, 35742549198872617291353508656626642567, 359334085968622831041960188598043661065388726959079837 ... ?

Usually when they jump right from small numbers into enormous numbers it means that the series is a list of numbers with an unusual property.
Spoiler:

EDIT:
Spoiler:

[SIGPIC][/SIGPIC]
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• Registered User regular
edited April 2008
Absurdist wrote: »
Apothe0sis wrote: »
This is most assuredly not my series, and math geeks will probably see why this is funny.

2, 5, 877, 27644437, 35742549198872617291353508656626642567, 359334085968622831041960188598043661065388726959079837 ... ?

Usually when they jump right from small numbers into enormous numbers it means that the series is a list of numbers with an unusual property.
Spoiler:
You probably have to recognise the series. It's concievably solvable, but you'd have to have an Erdos number of 0 or something to do so.

REAL SPOILER
Spoiler:

Provide sample data to the Traitor project here || What is Traitor?
SODOMISE INTOLERANCE
Tide goes in. Tide goes out.
·
• Registered User
edited April 2008
Apothe0sis wrote: »
Absurdist wrote: »
Apothe0sis wrote: »
This is most assuredly not my series, and math geeks will probably see why this is funny.

2, 5, 877, 27644437, 35742549198872617291353508656626642567, 359334085968622831041960188598043661065388726959079837 ... ?

Usually when they jump right from small numbers into enormous numbers it means that the series is a list of numbers with an unusual property.
Spoiler:
You probably have to recognise the series. It's concievably solvable, but you'd have to have an Erdos number of 0 or something to do so.

REAL SPOILER
Spoiler:

I'm just happy that I know what an Erdos number is.

[SIGPIC][/SIGPIC]
·
• Registered User regular
edited April 2008
Absurdist wrote: »
Nobody can solve my puzzlers. Do I win something?

10, 0, 3, 20, 16, 51, 32, 67, 74, ?
Spoiler:

Absurdist wrote: »

1, 4, 7, 12, 15, 18, 21, 24, 27, ?
Spoiler:

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• Registered User regular
edited April 2008
Here is a fairly easy one.

If my three were a four, and my one where a three, what I am would be nine less than half what I'd be.
I am a three digit, whole number. What am I?

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• Registered User regular
edited April 2008
Suckafish wrote: »
Here is a fairly easy one.

If my three were a four, and my one where a three, what I am would be nine less than half what I'd be.
I am a three digit, whole number. What am I?
Spoiler:

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• Registered User regular
edited April 2008
It has multiple answers, doesn't it?

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• Registered User regular
edited April 2008
Suckafish wrote: »
Absurdist wrote: »
Nobody can solve my puzzlers. Do I win something?

10, 0, 3, 20, 16, 51, 32, 67, 74, ?
Spoiler:

Absurdist wrote: »

1, 4, 7, 12, 15, 18, 21, 24, 27, ?
Spoiler:
And either way, it's not a math puzzle.

It's an easy game to hate
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• Registered User regular
edited April 2008
Absurdist wrote: »
Apothe0sis wrote: »
Absurdist wrote: »
Apothe0sis wrote: »
This is most assuredly not my series, and math geeks will probably see why this is funny.

2, 5, 877, 27644437, 35742549198872617291353508656626642567, 359334085968622831041960188598043661065388726959079837 ... ?

Usually when they jump right from small numbers into enormous numbers it means that the series is a list of numbers with an unusual property.
Spoiler:
You probably have to recognise the series. It's concievably solvable, but you'd have to have an Erdos number of 0 or something to do so.

REAL SPOILER
Spoiler:

I'm just happy that I know what an Erdos number is.

Speaking of, wouldn't anyone with an Erdos number of 0 be dead, and thus unable to solve the problem?

Burnage wrote:
FWD is very good at this game.
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• Registered User regular
edited April 2008
Scooter wrote: »
It has multiple answers, doesn't it?
Not for 3 digits, as far as I can tell.

More complex explanation:
Spoiler:

Also, Paul Erdös is the only person with an Erdös number of 0, and he is dead. There are a lot of people still alive with an Erdös number of 1, however. And Natalie Portman has an Erdös number of 5.

It's an easy game to hate
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• Registered User regular
edited April 2008
Spoiler:

·