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Thread: Cross-number Puzzle (3x3)

  1. #1
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    Cross-number Puzzle (3x3)

    hello everyone,

    kindly solve this puzzle. . .its gving me a headache

    P.S.: this is all that i have on this puzzle

    Cross-number Puzzle (3x3)-quiz.jpg
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  2. #2
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    Re: Cross-number Puzzle (3x3)

    Does one digit go in each rectangle?
    If so, what is the difference between "1 across" and "2 across"?
    Thanks from DenisB and topsquark
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  3. #3
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    Re: Cross-number Puzzle (3x3)

    You need instructions for what they mean. Without instructions for how to interpret, the problem is meaningless. What does "1 down" mean? What inputs are possible in each box? Is it talking about the measurements of the "squares"? Is it talking about sums of digits?
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  4. #4
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    Re: Cross-number Puzzle (3x3)

    Unfortunately, there are no accompanying instructions . . .
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  5. #5
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    Re: Cross-number Puzzle (3x3)

    Then, unfortunately, it is a meaningless puzzle unless someone wants to make up instructions for it.

    For example, suppose the boxes contain the digits $a,b,c,d,e,f,g,h,i$ like this:

    $\begin{bmatrix}a & b & c \\ d & e & f \\ g & h & i\end{bmatrix}$

    And we interpret the instructions as:

    $100a+10b+c = (100a+10d+g)^2$
    $100d+10e+f = \dfrac{9}{4}(100a+10d+g)^2$
    $100g+10h+i = (10f+i)^2$
    $7|(10b+c)$

    We find that we must have $a=0,b=4,c=9,d=0,g=7$. But, this gives a non-integer for the second equation. So, this is not a valid set of instructions.

    How about $a,...,i$ are positive integers? And we have:

    $a+b+c = (a+d+g)^2$
    $d+e+f = \dfrac{9}{4}(a+d+g)^2 = \dfrac{9}{4}(a+b+c)$
    $g+h+i = (f+i)^2$
    $7|(b+c)$

    Now, we can pretty much have any integers we want for at least 3 or 4 of the variables and it will still solve the puzzle.

    So, we need enough restrictions that we find a distinct solution and enough freedom that at least one solution exists. Without the proper instructions, it is unlikely that we will ever find a solution.
    Last edited by SlipEternal; Jul 11th 2017 at 08:40 AM.
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