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Really challenging puzzle

  1. Nov 20, 2005 #1
    Last edited: Nov 20, 2005
  2. jcsd
  3. Nov 21, 2005 #2
    very difficult :D
  4. Nov 21, 2005 #3
    The missing number is 27
    Add all the numbers = 757
    And this is on a quare shown
    so 784 = 27* 27
  5. Nov 21, 2005 #4
    You make some assumption, I think. :)
  6. Nov 21, 2005 #5


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    27*27 = 729, not 784.

    28*28 = 784.

    However, you could also say that the area of the missing piece is 84, so that the total area is 841 (=29*29).
  7. Nov 21, 2005 #6
    sorry Mr. ahrkron,
    You are right......it could be 28 or 84......
  8. Nov 21, 2005 #7
    It seems like it is just to confuse you with the shapes within the square. I was able to set the square up into a grid. With the grid( I may be pulling at straws here) it seems like the answer would be 45.
  9. Nov 29, 2005 #8
    This puzzle was approached on this forum.

    http://www.scienceforums.net/forums/showthread.php?t=14658 [Broken]

    I will bet any money that they have no clue ether.

    And actually it’s straight forward “honest” puzzle.
    Last edited by a moderator: May 2, 2017
  10. Nov 30, 2005 #9


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    or 143 (30^2-757)
    or 204 (31...)
    or 267 (32...)

    Sorry, no dice. These problems have unambiguous answers.
  11. Dec 2, 2005 #10
    My guess is the numbers are related to the angles that the individual shapes contain. A different value for 90 degrees, more than 90, or less than 90.

    Possibly the number is then subtracted from or added to a constant.

  12. Dec 2, 2005 #11


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    It think the shapes are arbitrary. I think the only correlation is the numbers and their neighbours. There are some tantalizing correlations when factoring the numbers, but nothing jumps out.
  13. Dec 3, 2005 #12
    Ok here is the approach I took. I have too many Math finals coming up anyways so I won't be solving this myself. Anyone who knows linear algebra is welcome to try this, I have no clue if it will work. But here it goes:

    After playing with the shapes and sums of numbers I realized that the numbers don't appear to describe the shapes. So I took the following outlook on the situation. Assume that the shapes represent provinces. Each province produces the same amount of rice each year, no matter its size.

    This means that if the barriers changed and the provinces moved around, that, the amount of rice produced would stay the same. I then realized that the shapes can basically be anything you want them to be, because they aren't representative of the numbers (as long as there are 16 of them).

    I decided to redraw the big square into 16 equal squares (since there are 16 shapes). I put the numbers randomly into any square (imagine the provinces fighting over land and moving around but still producing the same amount of rice.

    Then it dawned on me that if x were to be the number behind the question mark, r were to be the amount of (lets say) rice prduced by all provinces including x, and y would be some value that is a common sum within the rows/columns of my grid, then solving the puzzle is just a matter of finding three equations in three variables (with no two equations alike).

    I took [r = 757 + x], [r - x = 3y + (y - x)], and I haven't actually figured out the equation relating y and r. I'm thinking along the lines of [4r = y/4].
    However, if the relation [y] between the numbers is incorrect, then the answer you get from solving the system would be correct but not the one they are looking for. Though if you can actually find a [y] variable which every row/column sum to, I'd say you're on the right track.

    http://i7.photobucket.com/albums/y300/medrawuc/squarepuzzlesolution.jpg" [Broken]

    Anyways, give it a try if you are good at this stuff, my brain will be overloaded from Math in a few days anyways so I'm gonna lay off of it for now. :zzz:
    Last edited by a moderator: May 2, 2017
  14. Dec 4, 2005 #13
    i have my answer, using your approach:

    r = 804
    y = 201
    x = 47

    here is the matrix i made:

    74 66 49 57
    42 33 62 41
    55 42 47 56
    30 60 43 47
  15. Dec 4, 2005 #14
    Thanks, I was really hoping someone would pick that up. Does anyone know if it's correct?
  16. Dec 4, 2005 #15
    Isn't [r - x = 3y + (y - x)] just r=4y?

    So if 47 was a valid solution then so would 51 be. You'd just have to change the values of r and y.


    Or am I misunderstanding you?
  17. Dec 4, 2005 #16
    Yes that's the simplified formula, but I wanted to be clear on how that formula came about, hence the sort of redundant equation.

    As for changing [y] and [r]: Then its just a random answer. My approach was based on the idea that all the columns add to a certain value, that value, as we now see, is 201. If you made [y] 202 then 3 of the columns would add to 201 and the last one would add to 202. Thus defeating the entire purpose of the [y] variable.
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