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Posers and Puzzles

Posers and Puzzles

  1. Standard member clandarkfire
    Grammar Nazi
    19 Sep '09 06:16 / 2 edits
    It is possible to place positive integers into the twenty-one vacant squaress of the 5x5 square shown below so that the numbers in each row and column form an arithmetic sequence.

    *.............. *.............. *.............. *.............. *
    *.............. 74............ *.............. *.............. *
    *.............. *.............. *.............. *.............. 186
    *.............. *.............. 103.......... *.............. *
    0.............. *.............. *.............. *.............. *


    The periods are only there so it lines up nicely. I hope you guys get what I'm asking
    replace each asterix with a number and explain how you did it.
  2. 19 Sep '09 14:59
    052...082...112...142...172
    039...074...109...144...179
    026...066...106...146...186
    013...058...103...148...193
    000...050...100...150...200

    The key was mod 12 arithmetic as far as it would go, trying out relatively few possibilities.

    (Trial and error probably would have worked too.)
  3. 19 Sep '09 15:06
    Originally posted by clandarkfire
    It is possible to place positive integers into the twenty-one vacant squaress of the 5x5 square shown below so that the numbers in each row and column form an arithmetic sequence.

    *.............. *.............. *.............. *.............. *
    *.............. 74............ *.............. *.............. *
    *.............. *.............. *...... ...[text shortened]... you guys get what I'm asking
    replace each asterix with a number and explain how you did it.
    There are 3 numbers adjacent to 0. call them a,b,c:

    b c
    0 a

    they determine all the other numbers, for example

    ... ... ... ... ...
    ... ... ... ... ...
    2b 2c-a ... ... ...
    b c 2c-b 3c-2b 4c-3b
    0 a 2a 3a 4a

    you get 3 equations with 3 unknowns.
  4. Standard member clandarkfire
    Grammar Nazi
    19 Sep '09 18:08
    Thank you for your answers, the one given is obviously correct. However, could you please try to explain how you got them? I don't really understand what you are saying.
  5. Standard member clandarkfire
    Grammar Nazi
    19 Sep '09 19:37
    Originally posted by geepamoogle
    052...082...112...142...172
    039...074...109...144...179
    026...066...106...146...186
    013...058...103...148...193
    000...050...100...150...200

    The key was mod 12 arithmetic as far as it would go, trying out relatively few possibilities.

    (Trial and error probably would have worked too.)
    in other words, how did you use mod 12 arithmetics to solve this problem?
  6. 20 Sep '09 02:42
    Well, I started by finding the remainder of each of the given numbers when divided by 12, then deducing what the possibilities were for reamining slots.

    A good starting place was that the numbers 3 up and 3 over had to be divisible by 3, and the ones 4 up and 4 over divisible by 4. I'd have to work it out again to remember where I went from there, but the general idea was to eliminate the remainders which were impossible.

    Eventually, I was able to determine that the number above the zero was of the form 12x + 1.

    In examining the problem later, I have found a more useful pattern which can be used alongside multiple equation algebra to answer this sort of question.

    Be it noted here, that the above solution is the only solution possible, even if you allowed negative numbers, fractions, and real numbers.

    So now, here is a slightly modified form.

    ????...????...????...????...????
    ????...074....????...????...????
    ????...????...????...????...186
    ????...????...103...????...????
    040...????...????...????...????
  7. Standard member clandarkfire
    Grammar Nazi
    26 Sep '09 03:28
    About a week later, here is the answer: Its actually incredibly easy once you figure out how to do it.

    028...076...124...172...220
    031...074...117...160...203
    034...072...110...148...186
    037...070...103...136...169
    040...068...096...124...152
  8. Standard member TheMaster37
    Kupikupopo!
    28 Sep '09 11:50
    Originally posted by geepamoogle
    Well, I started by finding the remainder of each of the given numbers when divided by 12
    Why 12?
  9. 01 Oct '09 22:32 / 1 edit
    Originally posted by TheMaster37
    Why 12?
    Since each sequence is sequential in nature, and since I only have to deal with at most a difference of 4 terms in any sequence, I only need a number divisible by 1, 2, 3 and 4. 12 is the LCM. Alternately, one could track modulo 3 and modulo 4, since it has the same overall result. Apologies if this approach doesn't make any sense. I explain it poorly, although it makes perfect sense to me.

    However, examining the problem afterwards resulted in a discovery of a much more direct and efficient way of solving it.

    Examine the differences in each sequence in the grid and you should notice something interesting which proves very useful when coupled with multiple equation solving techniques.