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What's the most interesting thing you have learned this year in maths? Watch

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    (Original post by Kvothe the arcane)
    Solution to the Wason selection task.

    You are shown a set of four cards placed on a table, each of which has a number on one side and a colored patch on the other side. The visible faces of the cards show 3, 8, red and brown. Which card(s) must you turn over in order to test the truth of the proposition that if a card shows an even number on one face, then its opposite face is red?

    I thought the answer easy and intuitive but it turns out my reasoning was flawed.
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    Isn't the answer just 8 and brown?
    If from what you have said the number being even is sufficient to conclude that the other face is red, but the face being red is not sufficient to conclude that the number is even as the statement does not say that red belongs exclusively to even numbers.
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    (Original post by Aph)
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    Isn't the answer just 8 and brown?
    If from what you have said the number being even is sufficient to conclude that the other face is red, but the face being red is not sufficient to conclude that the number is even as the statement does not say that red belongs exclusively to even numbers.
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    Yes
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    Looking into quarternions
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    5 Pirates Problem

    5 pirates of different ages have a treasure of 100 gold coins. On their ship, they decide to split the coins using this scheme: The oldest pirate proposes how to share the coins, and ALL pirates (including the oldest) vote for or against it. If 50% or more of the pirates vote for it, then the coins will be shared that way. Otherwise, the pirate proposing the scheme will be thrown overboard, and the process is repeated with the pirates that remain.As pirates tend to be a bloodthirsty bunch, if a pirate would get the same number of coins if he voted for or against a proposal, he will vote against so that the pirate who proposed the plan will be thrown overboard.Assuming that all 5 pirates are intelligent, rational, greedy, and do not wish to die, (and are rather good at math for pirates) what will happen?

    Solutions?

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    My solution:

    Calling the pirates A (oldest), B, C, D and E(youngest)

    A: 98, B: 0, C: 1, D: 0, E: 1
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    (Original post by offhegoes)
    5 Pirates Problem

    5 pirates of different ages have a treasure of 100 gold coins. On their ship, they decide to split the coins using this scheme: The oldest pirate proposes how to share the coins, and ALL pirates (including the oldest) vote for or against it. If 50% or more of the pirates vote for it, then the coins will be shared that way. Otherwise, the pirate proposing the scheme will be thrown overboard, and the process is repeated with the pirates that remain.As pirates tend to be a bloodthirsty bunch, if a pirate would get the same number of coins if he voted for or against a proposal, he will vote against so that the pirate who proposed the plan will be thrown overboard.Assuming that all 5 pirates are intelligent, rational, greedy, and do not wish to die, (and are rather good at math for pirates) what will happen?

    Solutions?
    Spoiler:
    Show
    My solution:

    Calling the pirates A (oldest), B, C, D and E(youngest)

    A: 98, B: 0, C: 1, D: 0, E: 1
    I enjoy the intuition-busting nature of this one. I guess I'll do some explaining; it's late and I'm bored
    Spoiler:
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    As you've done, label the pirates A through E by age.

    Let's examine what would happen with just D and E left. D could clearly take 100 coins for himself and vote for his proposal. That is, the following arrangement necessarily occurs:
    D: 100
    E: 0

    So what if just C, D and E are left? If C takes 100 coins for himself, the other two will get 0, and then they will surely vote to kill him as they will either do better by doing so or at least not do any worse. But we note that if E is afforded any coins then E will prefer to keep C alive, or else E will lose money, since the above scenario follows. So C can simply take 99 coins for himself and give 1 to E, then C and E will both vote for C to live, i.e. we have
    C: 99
    D: 0
    E: 1

    Next, B,C,D and E. We have a very similar case. B cannot survive by taking 100 coins, but if D is given any money, D stands to lose from voting against the proposal, as the above arrangement, in which D has 0 coins, occurs. So B can keep 99 coins for himself and give 1 to D, to ensure that he gets 50% of the votes, namely his own and D's, yielding
    B: 99
    C: 0
    D: 1
    E: 0

    Finally, we consider all five pirates alive and well. Here A can at most have 98 coins, or else at least three pirates will have 0 coins, and so they will vote against him. But note, in similar manner to before, that should C and E be given any money they will vote in A's favour to avoid the above scenario in which they have no coins. So A simply gives 1 coin to C, 1 coin to E, and keeps the rest for himself. This gives the final answer
    A: 98
    B: 0
    C: 1
    D: 0
    E: 1

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    Prepared a presentation for the Naviar Stokes equations. Part of it was a simple gas flow derivation that I'd found on that internet, and it was quite amazing to see simple physical variables be magically transformed into intuitive differential equations
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    The monkeys and coconuts problem
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    Are there too many people showing off in this thread?
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    (Original post by M14B)
    Are there too many people showing off in this thread?
    If by showing off you mean answering the question then yeah


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    (Original post by M14B)
    Are there too many people showing off in this thread?
    I'm not sure if 'showing off' is the best way to describe it. It is maths after all. No-one is getting laid off the back of this thread.
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    (Original post by Redcoats)
    Is that true for any x,y?


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    (Original post by Redcoats)
    Ive never come across mod(a,b) notation. What does this mean.


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    (Original post by physicsmaths)
    Ive never come across mod(a,b) notation. What does this mean.


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    It's called Tupper's Self Referential Formula (also called 'the everything formula')

    Mod(a,b) is where a is the number for which you want to find the remainder and b is the divisor.

    Check out this video on it - you'll be glad you did; what the formula can do is insane!

    https://www.youtube.com/watch?v=_s5RFgd59ao
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    (Original post by Redcoats)
    It's called Tupper's Self Referential Formula (also called 'the everything formula')

    Mod(a,b) is where a is the number for which you want to find the remainder and b is the divisor.

    Check out this video on it - you'll be glad you did; what the formula can do is insane!

    https://www.youtube.com/watch?v=_s5RFgd59ao
    That the one they used to make the batman graph?


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    (Original post by drandy76)
    That the one they used to make the batman graph?


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    No, it does way better than that!
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    (Original post by Redcoats)
    No, it does way better than that!
    Are you implying that anything in this world is greater than the dark knight?
    (Like what?)


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    2+2=5
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    Calculating whether different stars can float in a a bathtub and also binomial expansion.
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    (Original post by Redcoats)
    It's called Tupper's Self Referential Formula (also called 'the everything formula')

    Mod(a,b) is where a is the number for which you want to find the remainder and b is the divisor.

    Check out this video on it - you'll be glad you did; what the formula can do is insane!

    https://www.youtube.com/watch?v=_s5RFgd59ao
    Its writte a Z_b or a (mod b) didn't know skme people write it like that.


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