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[eHalcyon]

Not so much a logic problem, but a fun problem anyway:

Using each of the numbers 1, 3, 4 and 6, each exactly once, as well as any number of +, -. x and ÷ as well as brackets, make the number 24.

I think there's only one solution, and it took me at least 40 minutes to find

6/(1-3/4)

(14-6)x3

I think combining 1 and 4 to make 14 is not allowed within the problem parameters.


This exercise can actually be played as a card game with a regular deck of cards.  I am terrible at this game.

[ConMan]

http://en.wikipedia.org/wiki/Factorial#Definition

http://www.zero-factorial.com/whatis.html

It depends on how you define 0!, and it turns out that the most consistent, most logical, and most convenient (if not most immediately intuitive) definition of 0! is 1

I define factorial as the Gamma function, translated by 1 and restricted to integers, from which 0! = 1 is an automatic consequence.

[Drab Emordnilap]

I define factorial as the Gamma function, translated by 1 and restricted to integers, from which 0! = 1 is an automatic consequence.

I define factorial as me yelling at numbers until they're the numbers I want them to be.

[dondon151]

This exercise can actually be played as a card game with a regular deck of cards.  I am terrible at this game.

The one thing that I can't do with this game is working with fractions.

[Tables]

Not so much a logic problem, but a fun problem anyway:

Using each of the numbers 1, 3, 4 and 6, each exactly once, as well as any number of +, -. x and ÷ as well as brackets, make the number 24.

I think there's only one solution, and it took me at least 40 minutes to find

6/(1-3/4)

(14-6)x3

I specifically say numbers, not digits, to avoid this kind of thing. Still I didn't realise that way was also possible.

Also surprised eHalc got the solution so quickly, all things considered...

[eHalcyon]

Not so much a logic problem, but a fun problem anyway:

Using each of the numbers 1, 3, 4 and 6, each exactly once, as well as any number of +, -. x and ÷ as well as brackets, make the number 24.

I think there's only one solution, and it took me at least 40 minutes to find

6/(1-3/4)

(14-6)x3

I specifically say numbers, not digits, to avoid this kind of thing. Still I didn't realise that way was also possible.

Also surprised eHalc got the solution so quickly, all things considered...

I've seen the same type of problem before, where the trick is to use fractions in that manner.  And a friend asked that kind of problem of me only about a month ago, so it was kind of fresh in my memory.

[Axxle]

Just because its true for one set of numbers, doesn't make it true for 0!

1337!/1336! = 1337

58008!/58007! = 58008

3720!/3719! = 3720

0!/(-1)! =

[ConMan]

Just because its true for one set of numbers, doesn't make it true for 0!

1337!/1336! = 1337

58008!/58007! = 58008

3720!/3719! = 3720

0!/(-1)! =

Hence why factorial is defined only on the natural numbers, and the gamma function has essential singularities at the negative naturals.

[Axxle]

Just because its true for one set of numbers, doesn't make it true for 0!

1337!/1336! = 1337

58008!/58007! = 58008

3720!/3719! = 3720

0!/(-1)! =

Hence why factorial is defined only on the natural numbers

So why is it defined for zero?

edit: I guess 0 is a natural number in some circles

[ConMan]

Just because its true for one set of numbers, doesn't make it true for 0!

1337!/1336! = 1337

58008!/58007! = 58008

3720!/3719! = 3720

0!/(-1)! =

Hence why factorial is defined only on the natural numbers

So why is it defined for zero?

edit: I guess 0 is a natural number in some circles

Yeah, I think we've had this discussion on this board already (although I didn't chime in), but I count natural numbers as the set {0, 1, 2, ...}. So n! is defined for n = 0, 1, 2, ... only, whereas Gamma(x) is defined for all complex x except x = 0, -1, -2, ... But Gamma(n+1) = n! which is why there's a difference between where they work.

[DStu]

Just because its true for one set of numbers, doesn't make it true for 0!

1337!/1336! = 1337

58008!/58007! = 58008

3720!/3719! = 3720

0!/(-1)! =

Hence why factorial is defined only on the natural numbers

So why is it defined for zero?

edit: I guess 0 is a natural number in some circles

I don't think it is some ideological what is a natural number thing.
It is just that the factorial is probably most often used as "how many ways can you order a n element set" (including n chose k), and there the answer for 0 is 1. And you need to define 0! tO have the usual definnition dor n chose 0 and n chose n work.  And beside that mathematicians like to define things as broad as possible as long as there is some "natural" way to do it, and here it fits nicely into the recursive definition.

[Axxle]

Ok, that makes more sense.

[theory]

Back on topic: relevant SMBC:

[pacovf]

Avid reader, huh? That one's fresh.

[eHalcyon]

Don't forget the bonus:

[DStu]

I don't want to be impatient, but how does the puzzle with the infinite hats work?

[qmech]

I'll do the version where the people are labelled 1, 2, 3, ... and everybody knows which position they are in.  Generalising to other cases is still an interesting problem.  I repeat my warning that if you're a mathematics student you should resist the temptation to read the answer.

Define an equivalence relation on the set of hat colourings by saying that two colourings are equivalent if they differ in only finitely many places.  Choose a representative R(E) for each equivalence class E.  Everyone can see all but finitely many hats, so can determine which equivalence class E they are in.  Have everybody guess the colour their hat would be in the colouring R(E).  The actual colouring is equivalent to R(E), so only finitely many people will guess wrong.

For those who aren't familiar with them, equivalence relations aren't scary at all.  They're just a convenient way to describe breaking a set into pieces: two objects are in the same piece if and only if they are equivalent.

The possible generalisations are to the cases where people don't know which number they are, and where the places are indexed by Z instead of N (people can still only see in the "positive" direction).  I'm not claiming that all of these are doable.

[DStu]

Define an equivalence relation on the set of hat colourings by saying that two colourings are equivalent if they differ in only finitely many places.  Choose a representative R(E) for each equivalence class E.  Everyone can see all but finitely many hats, so can determine which equivalence class E they are in.  Have everybody guess the colour their hat would be in the colouring R(E).  The actual colouring is equivalent to R(E), so only finitely many people will guess wrong.

Hmm, that seems to simple to work.  But somehow...

[Asper]

How about a Puzzle without infinity, hats, or untimely death?

You have 12 bowls. All look the same. One is either heavier or lighter than the others, but the remaining ones weight the same.
All you have is a pair of scales, so you can compare any two sets of bowls on their weight. You may only do this 3 times, though.
Your task is to find out which bowl is different, and whether it's lighter or heavier than the others, or to prove this is impossible.

[Axxle]

How about a Puzzle without infinity, hats, or untimely death?

You have 12 bowls. All look the same. One is either heavier or lighter than the others, but the remaining ones weight the same.
All you have is a pair of scales, so you can compare any two sets of bowls on their weight. You may only do this 3 times, though.
Your task is to find out which bowl is different, and whether it's lighter or heavier than the others, or to prove this is impossible.

2^3 < 12
it's impossible. (I think I figured out a way to be accurate 66% of the time and have a 50/50 shot the last 33% of the time, but it's not 100%.

[DStu]

How about a Puzzle without infinity, hats, or untimely death?

Na, I don't like that.  How about:

You have 12 hats. All look the same. One is either heavier or lighter than the others, but the remaining ones weight the same.
All you have is a pair of scales, so you can compare any two sets of hats on their weight. You may only do this 3 times, though.
Your task is to find out which hat is different, and whether it's lighter or heavier than the others, or to prove this is impossible.

[Ozle]

Pretty sure it is doable.....

[DStu]

2^3 < 12
it's impossible. (I think I figured out a way to be accurate 66% of the time and have a 50/50 shot the last 33% of the time, but it's not 100%.

I'm not quite sure at the moment, maybe I only know the variante where you know if the one hat weights more or less, but the trick is that

3ˆ3 > 12

[Asper]

How about a Puzzle without infinity, hats, or untimely death?

You have 12 bowls. All look the same. One is either heavier or lighter than the others, but the remaining ones weight the same.
All you have is a pair of scales, so you can compare any two sets of bowls on their weight. You may only do this 3 times, though.
Your task is to find out which bowl is different, and whether it's lighter or heavier than the others, or to prove this is impossible.

2^3 < 12
it's impossible. (I think I figured out a way to be accurate 66% of the time and have a 50/50 shot the last 33% of the time, but it's not 100%.

Fun fact: My Math teacher tried to prove it's impossible when i was in school. Your proof is wrong.

[Asper]

How about a Puzzle without infinity, hats, or untimely death?

Na, I don't like that.  How about:

You have 12 hats. All look the same. One is either heavier or lighter than the others, but the remaining ones weight the same.
All you have is a pair of scales, so you can compare any two sets of hats on their weight. You may only do this 3 times, though.
Your task is to find out which hat is different, and whether it's lighter or heavier than the others, or to prove this is impossible.

Don't forget to mention the king will kill you if you can't do it.