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Olympiad problem

  1. Sep 27, 2011 #1
    Q. Sixty four squares of a chess board are filled with positive integers one on each in such a way that each integer is the average of the integers on the neighboring squares. (Two squares are neighbours if they share a common edge or vertex. Thus a square can have 8,5 or 3 neighbours depending on its position.) Show that all the sixty four entries are in fact equal.

    How does one begin this problem? I initially thought of assuming the contrary(that all the numbers are distinct) and then tried to obtain some ridiculous conclusion. But all I got were equations with innmuerable unknowns! Can anyone please give me a hint on starting the problem? Please DO NOT post the entire solution, as I would like to solve this on my own. Thanks! :D
     
  2. jcsd
  3. Sep 27, 2011 #2

    Mark44

    Staff: Mentor

    I haven't worked this problem, but I think your plan of assuming the contrary is a good one. Keep in mind however, that the opposite (logical negation) of "all 64 entries are equal" is NOT "all 64 entries are distinct."

    The opposite is "not (all 64 entries are distinct)." Another way of saying this is that two or more of the entries are different.
     
  4. Sep 27, 2011 #3

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    Pick n to be the minimum of the integers on the square. Now what? This actually seems pretty easy for an olympiad problem.
     
    Last edited: Sep 27, 2011
  5. Sep 28, 2011 #4
    Since this is a finite set of integers, then there must be a greatest integer in the set, call it n. Since n is an average of its surrounding squares, it cannot be strictly greater than every surrounding square, therefore it must be less than or equal to every surrounding square. However since it is the greatest integer in the set, the set must have only 1 integer element, and all squares are equal.
     
  6. Sep 28, 2011 #5
    Ah, so that's how it's done...thanks a lot guys! It seems I'm not Olympiad material after all :D...I was so sure that I could do this on my own, and yet I goofed...
     
  7. Sep 28, 2011 #6
    I've got another one: Find the least number that ends with 7, so that when you put the 7 at the beginning(and removing the 7 at the end), you get a number 5 times larger as that of the original one.
     
  8. Sep 28, 2011 #7
    I'd start by noticing my ten's digit in the original number must be a 0 or a 5 (in order for it to be divisible by 5).

    Actually, on second thought, it must be a 5, and can't be a zero.

    Also, the leading digit has to be a 1 since if it is a 2 or more, multiplying it by 5 won't give a leading digit of 7.
     
    Last edited: Sep 28, 2011
  9. Sep 28, 2011 #8
    After thinking about it, it was actually easier than I thought. The last digit has to be 7 and the second to last has to be 5 (so the number is at least 57), so 5 x 7 =35 + 50 = 85, so the third to last digit has to be 8, and the number is at least 857. Continuing on in this fashion, then the number is 142857
     
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