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asnaseer

  • 4 years ago

Reposting puzzle since no one has solved it yet: You are allowed to use any number of mathematical operators or functions (e.g. +, -, *, /, sin, cos, tan, factorial, sqrt, powers, etc) EXCEPT floor and ceiling functions. The only constants that can appear in the solution are two digits - both of which must be '2'. Use these rules to make the number 5.

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  1. FoolForMath
    • 4 years ago
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    Can we use decimal point?

  2. asnaseer
    • 4 years ago
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    yes - looks like you may have found one of the solutions (I know 2 ways of doing this) :-)

  3. FoolForMath
    • 4 years ago
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    Well, don't you know I am the smart one? :P :D

  4. asnaseer
    • 4 years ago
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    of course I do :-)

  5. FoolForMath
    • 4 years ago
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    Anyways, I won't spoil the problem for others, but here is something related http://en.wikipedia.org/wiki/2_%2B_2_%3D_5

  6. asnaseer
    • 4 years ago
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    Wow - I guess you really do learn something new everyday - I was not aware of this - thx

  7. FoolForMath
    • 4 years ago
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    You are welcome :)

  8. asnaseer
    • 4 years ago
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    btw - there is a more /pure/ solution which does not involve using a decimal point

  9. FoolForMath
    • 4 years ago
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    I suppose we can't use variable substitution too, isn't ?

  10. asnaseer
    • 4 years ago
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    what do you mean @FoolForMath?

  11. Zed
    • 4 years ago
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    do the powers need to be 2 or anything?

  12. asnaseer
    • 4 years ago
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    remember the rules @Zed - you can ONLY have two constants in the resulting expressions and BOTH must be the digit '2'

  13. Zed
    • 4 years ago
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    ahh i see, read the question wrong

  14. FoolForMath
    • 4 years ago
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    Hm something like this: x=2; x^x + x^(x-x) = 5 :D wait that's with only one 2 :D

  15. asnaseer
    • 4 years ago
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    no sorry @FoolForMath - that is not allowed.

  16. asnaseer
    • 4 years ago
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    CLUE: the /purer/ solution involves right-angled triangles

  17. FoolForMath
    • 4 years ago
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    Hm then how about this one \( \csc^2(\cot^{-1}(2)) \) , Checking: http://www.wolframalpha.com/input/?i=Csc%5BArcCot%5B2%5D%5D%5E2

  18. asnaseer
    • 4 years ago
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    thats the one! - well done @FoolForMath can the others find the other solution?

  19. FoolForMath
    • 4 years ago
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    Thanks asnaseer :) Should I deleted it ?

  20. asnaseer
    • 4 years ago
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    5=cosec^2(arctan(2))

  21. Zed
    • 4 years ago
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    Well done FoolForMath

  22. asnaseer
    • 4 years ago
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    that is how I wrote it to avoid the "-1"

  23. asnaseer
    • 4 years ago
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    |dw:1325595329168:dw|

  24. asnaseer
    • 4 years ago
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    no need to delete your answer - lets see if anyone can find the other way of doing this

  25. asnaseer
    • 4 years ago
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    CLUE for other way: uses one decimal point, one squareroot and one power

  26. FoolForMath
    • 4 years ago
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    The other one is a bit hard in my opinion, I have seen this problem before sooo ;) anyways, the latex version of my solution \( \csc^2(\arccot(2)) \)

  27. asnaseer
    • 4 years ago
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    \(cosec^2(arccot(2))\)

  28. asnaseer
    • 4 years ago
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    I wrote arctan above which was wrong.

  29. FoolForMath
    • 4 years ago
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    btw hey your earlier post is not correct 5 \( \neq \)cosec^2(arctan(2))

  30. asnaseer
    • 4 years ago
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    snap!

  31. FoolForMath
    • 4 years ago
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    wait, arccot is defined in latex no ?

  32. FoolForMath
    • 4 years ago
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    See: \[ \arccot \arctan \]

  33. asnaseer
    • 4 years ago
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    hmmm...

  34. Zed
    • 4 years ago
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    okay i think this is right \[\sqrt{0.2^{-2}}=5\]

  35. asnaseer
    • 4 years ago
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    yup - you got it @Zed - well done!

  36. FoolForMath
    • 4 years ago
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    yep! you got that right :)

  37. Zed
    • 4 years ago
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    Yay! That was tricky :)

  38. asnaseer
    • 4 years ago
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    the only thing I don't like about this way is that sqrt(x) is really x^-1/2

  39. Zed
    • 4 years ago
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    yeah i agree

  40. asnaseer
    • 4 years ago
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    so it /feels/ like cheating - but nevertheless - well done @Zed

  41. FoolForMath
    • 4 years ago
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    Hero wants hard, lets give him a bit harder :D I just found two other ways of getting the same result, could anybody wanna try ?

  42. FoolForMath
    • 4 years ago
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    Guys if you are trying, then please note it is not a good one and the second is ugly lol, and there is not much of useful maths we can learn from those two.

  43. asnaseer
    • 4 years ago
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    no - I was just waiting for @Hero to return to take up the challenge :-)

  44. FoolForMath
    • 4 years ago
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    lol :D asnaseer, wanna try something from my sleeve ? ;)

  45. asnaseer
    • 4 years ago
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    gulp! - I know your reputation @FoolForMath - but lets for it - you gonna post to the left?

  46. asnaseer
    • 4 years ago
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    *lets GO for it

  47. FoolForMath
    • 4 years ago
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    yeah sure :D

  48. ixforrest
    • 4 years ago
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    ummm this is kind o really easy.. at first I thought it was hard but i realized i was thinking too much.... 2^2 +2^3 = (12)^2 = 144 /2 = 72 /2 =36 /2 =18/2 =9 √9 = 3+2 = 5

  49. ixforrest
    • 4 years ago
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    no need for big guns this was basic problem solving skills

  50. asnaseer
    • 4 years ago
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    you didn't read the rules properly - the ONLY constants allowed are two digits - both of which must be the digit 2. e.g. 2^2+2^3 uses three 2's and a 3 - breaking the rules.

  51. ixforrest
    • 4 years ago
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    eh it was one of three or four answers ive got more. threre is also something interesting that happens between radians, but let me guess the implied conversion of a radian as 180 counts as a number other than 2 ? I still dont believe this requires higher level math I will stab at this later for now I have a date w/ Mathematica 8 and a vat of coffee. Thanks for clarifying !

  52. asnaseer
    • 4 years ago
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    :-D - enjoy the coffee!

  53. ixforrest
    • 4 years ago
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    always!

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