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Featured B Four 4s puzzle

  1. Feb 7, 2017 #1
    Say you've got four 4s -- 4, 4, 4, 4 -- and you're allowed to place any normal math symbols around them. How many different numbers can you make? It's best to think of a number and then try to make it.

    The origination of the puzzle and some examples are in this video
     
  2. jcsd
  3. Feb 7, 2017 #2

    Student100

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    ##4444\int dx =4444x +C## I think that about covers it. o:)
     
  4. Feb 7, 2017 #3

    jedishrfu

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    Imagine what you could do with seven fours?

    You could order Pizza Hut pizza! (From an old pizza commercial)

    Nice video. I was waiting for Knuth arrow notation to come in but it never did.
     
  5. Feb 8, 2017 #4
    I think that wraps things up!
     
  6. Feb 8, 2017 #5

    martinbn

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    No, it doesn't. How do represent the number 10 for example?
     
  7. Feb 8, 2017 #6
    I think the log based proof was quite good for numberphile.

    Also it seems you are allowed to put the minus sign in front of the log function which would extend the combos into the negative numbers for reals
     
  8. Feb 8, 2017 #7
    Can anyone figure out Pi?
     
  9. Feb 8, 2017 #8
    0 could also be done with something log based like

    lg(4/4)=0
     
  10. Feb 8, 2017 #9

    I'm not quite certain if this is allowed by the problem such as using variable x, such as using standard symbols and four 4s as numbers...
    But Pi can be gotten in certain ways

    ##lim_{x-> -1^{+}} =cos^{-1}(x)##

    I'm not sure if cos^{-1} was defined exactluy at the value of (-1)

    so that might be something like cos^{-4/4}(-4/4)= pi

    But limit should still exist. I suck at trig.

    that is in radians of course
     
  11. Feb 8, 2017 #10
    This is indefinite integral, i.e. function ...

    For any real number x: x = C/4444 . Just pick a C = x•4444 ...

    Or x = C/(4+4+4+4), or x = C/(4•4•4•4) ...

    or even (different form) x = C(4444) [C=x/4444], etc., or use other operations or operators ...

    Thus trivial, unless you put more constraints in the problem.
     
  12. Feb 8, 2017 #11

    Student100

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    It wasn't meant to be taken seriously.

    As for ten, x = 0, C = 10. There you go.
     
  13. Feb 8, 2017 #12
    Assigning variables is kind of cheating :D
     
    Last edited: Feb 8, 2017
  14. Feb 8, 2017 #13

    mfb

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    $$\pi = \sqrt{4}~ \arccos (\sqrt{4\cdot 4}-4) = \arccos \left(4-4-\frac 4 4 \right)$$
    Similar expressions exist with the arcsine.
    $$e=\exp\left(4-4+\frac 4 4 \right)$$
    Well...
     
  15. Feb 8, 2017 #14

    PAllen

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    Let x=0, C=10
     
  16. Feb 8, 2017 #15
    It would appear so (and mine above could also be considered partially humour), but isn't solving an equation like inverting an operator or a function? Thus cosistent with:

    And it is not limited by the video, or the original post [but since the OP sais now so we can now perhaps put it as a restriction ... (?)].

    For example (aside/[or plus] the fact that functions can be considered as operarors, and operators are essentially functions):

    If x is any real number (say π), and f any invertable real function (to avoid regular or perplexity form plain equations), and it happens that a number A which is directly/easily or else consructed by the 4 4s (say 4444) is related to x as A = f(x), then x = f-1(A)
    [e.g. x (or e.g. π) = f-1(4444), and that's true for any [invertable] function f (providing it satisfies A = f(x) ) (e.g. a•x, sin, cos, exp, ln, log etc.) - just pick an appropriate one and be my guest! ...]

    That's what I essentially did above, e.g.
    where f(x) = (1/C)•x, A = 4444, and mfb did essentially the same with cos and arccos
    since arccos = cos-1 ...

    So are we joking or serious? I am a bit confused.

    My opinion is that the problem on the video etc. is either trivial or one would have to put explicitly more restrictions, namely to the basic functions and operations only, with no arbitrary constants (other than the 4 4s and their obvious basic step derivatives).
    [In other words functions e.g. like a•x, or a•cosx ... are not allowed, unless a=1 ... etc., otherwise one can do anything - it's trivial]

    So the problem is not the variables but the arbitrary constants.
     
    Last edited: Feb 8, 2017
  17. Feb 8, 2017 #16

    mfb

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    I used common mathematical functions.
    I didn't have to make up a function just for this exercise.
     
  18. Feb 8, 2017 #17
    I know. Your's is ok

    As long as the OP accepts inverse functions.

    Mine has arbitrary constants, so may be not allowed, but the restrictions should be made explicit.
     
  19. Feb 8, 2017 #18

    mfb

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    There is nothing inverse about it. The arccosine is a regular function (in more than one way). The cosine (restricted to [0,pi]) is the inverse function of the arccosine. And now?
     
  20. Feb 8, 2017 #19
    But one has to define explicitly which functions are allowed to build from, and if inversing is allowed. (+see my edited posts above about arbitrariness of constants) But in any case I think your's is ok. I just wasn't sure if you were serious or not.
     
    Last edited: Feb 8, 2017
  21. Feb 8, 2017 #20

    Student100

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    Trying to come up with a simple way to get closer...

    Using factorial's and decimals you can get approximately close:

    $$\sum_{n=1}^{n=4} .4! = 3.54905... $$

    What about...

    $$ .04! \times 4 - .4 -.4 = 3.113752 $$
     
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