Posers and Puzzles

Posers and Puzzles

  1. Joined
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    31 Aug '14 19:18
    Earlier in the year I gave four problems from a book I bought. I just bought another book from the same store. I give two problems from it.

    i) A square piece of paper has numbers written at the corners. Specifically, the numbers going clockwise from upper left are 1, 2, 3, 10. Some kind soul has made copies--effectively an unlimited number of copies--of this item available to you. Can you choose some number of the squares so that, with proper rotation of each square in the stack (by some multiple of 90 degrees and keeping the numbered sides of the squares facing up), the sum of the numbers in each of the four corners is 67?

    ii) For what real number value of x is the xth root of x a maximum?
  2. Joined
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    31 Aug '14 20:43
    I should clarify, since I am not able to reproduce the book's diagram of the situation--

    The paper squares in (i) are stacked by you on top of one another such that edges line up neatly.
  3. Standard memberwolfgang59
    Mr. Wolf
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    01 Sep '14 03:33
    Originally posted by Paul Dirac II
    Earlier in the year I gave four problems from a book I bought. I just bought another book from the same store. I give two problems from it.

    i) A square piece of paper has numbers written at the corners. Specifically, the numbers going clockwise from upper left are 1, 2, 3, 10. Some kind soul has made copies--effectively an unlimited number of copies ...[text shortened]... 67?

    ii) For what real number value of x is the xth root of x a maximum?
    ii) Intuitive guess-work.
    Since 2 and 4 return the same result I'm guessing that 2<x<4

    And since Maths problems are elegant I'm guessing x=e or x=pi
    30 secs on a calculator tells me its not pi but possibly e.

    Am I close?
    😀
  4. Joined
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    01 Sep '14 04:561 edit
    Originally posted by wolfgang59
    ii) Intuitive guess-work.
    Since 2 and 4 return the same result I'm guessing that 2<x<4

    And since Maths problems are elegant I'm guessing x=e or x=pi
    30 secs on a calculator tells me its not pi but possibly e.

    Am I close?
    😀
    You are certainly on the right track, though neither e nor pi is the precise answer, and in fact your anticipated range is not where the answer falls.
  5. Joined
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    01 Sep '14 06:067 edits
    Upon further review...

    I am giving wolfgang credit for the right answer to (ii).

    The book's answer to "For what value of x is..." is "e^(1/e)."

    Goofing around on a calculator, I see that e^(1/e) = 1.444 approximately. Taking the 1.444th root of 1.444 yields 1.27, which is less than the square root of 2 (that root being about 1.414). So the book goofs. They should have said, "e is the value of x, because the function's value reaches a maximum of e^(1/e) for that value of x."

    It's a good thing we have the official's instant replay review rule going here. 😛

    The book does not show how to derive this, but attributes the result to Jacob Steiner. I will guess you start by taking the natural log of both sides of y = x^(1/x), and then derivatives set to zero to find where the maximum occurs, since the logarithm should max at the same place as the argument.
  6. Joined
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    01 Sep '14 07:131 edit
    Here's a method that gives the answer to (ii).

    Write x^(1/x) as e^[ln(x)/x].

    The derivative of e^u(x) is e^u(x) * du/dx.

    Taking the derivative of our function gives:
    e^[ln(x)/x] * {x^-2 - ln(x)/[x^2]}.

    Setting this to zero, and noting that the factor in front of the * cannot be zero, the second factor must vanish:

    x^-2 - ln(x)/[x^2] = 0.

    Multiplying through by x^2:
    1 - ln(x) = 0.

    So ln(x) = 1, meaning x=e.
  7. Standard memberwolfgang59
    Mr. Wolf
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    01 Sep '14 07:50
    Originally posted by Paul Dirac II
    I am giving wolfgang credit for the right answer to (ii).

    [/b]
    Thanks ... but not accepting it!

    My Sherlock Holmes approach was not very vigorous!!
  8. SubscriberPonderable
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    01 Sep '14 16:11
    Originally posted by Paul Dirac II
    Earlier in the year I gave four problems from a book I bought. I just bought another book from the same store. I give two problems from it.

    i) A square piece of paper has numbers written at the corners. Specifically, the numbers going clockwise from upper left are 1, 2, 3, 10. Some kind soul has made copies--effectively an unlimited number of copies ...[text shortened]... 67?

    ii) For what real number value of x is the xth root of x a maximum?
    i) would be trivial if the sum was 64. Since then you would stack 16 papers turning by 90 degrees after every four sheets.
    You obtain 10*4+3*4+2*4+1*4=64

    With 67 the problem oocurs that you would have to write down the possible permutations of the numbers giving 67, choose permutations whose hnumber of elemnts are at least 4, and then see if the permutations will fit the series...

    too much for me doing this.
  9. Joined
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    01 Sep '14 19:23
    Originally posted by Ponderable
    too much for me doing this.
    The book shows a remarkably simple way of deciding the answer. No exhausting computing is required!
  10. Standard memberwolfgang59
    Mr. Wolf
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    02 Sep '14 00:07
    Originally posted by Paul Dirac II
    Earlier in the year I gave four problems from a book I bought. I just bought another book from the same store. I give two problems from it.

    i) A square piece of paper has numbers written at the corners. Specifically, the numbers going clockwise from upper left are 1, 2, 3, 10. Some kind soul has made copies--effectively an unlimited number of copies ...[text shortened]... 67?

    ii) For what real number value of x is the xth root of x a maximum?
    1. For the sum of each corner to be 67 the total of all four corners must be 268. Since 268 is not a multiple of 16 (10+3+2+1) it is not possible.

    That seems far too easy. Have I misunderstood the problem?
  11. Joined
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    02 Sep '14 03:30
    Originally posted by wolfgang59
    1. For the sum of each corner to be 67 the total of all four corners must be 268. Since 268 is not a multiple of 16 (10+3+2+1) it is not possible.

    That seems far too easy. Have I misunderstood the problem?
    That's the right answer, and that is the recommended approach to it.

    No muss, no fuss. 😉
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