Wednesday, January 14, 2015

373: Product Puzzle 4

For the last couple of days, I've been asking you to create a new puzzle. Today, I'd like you to describe HOW to create a puzzle with a unique solution, or two solutions, or three. How did Don Steward create these and KNOW that they only had one solution?


1 comment:

  1. # A puzzle is defined:

    A "solution"

    A, B
    C, D

    exists such that

    W = A*B
    X = C*D
    Y = A*C
    Z = B*D

    where `W`, `X`, `Y`, `Z` are given.


    Let's define the "Solution Product" as the product of two rows or two columns.

    SP = W*X = Y*Z = A*B*C*D

    The Solution Product is constant,


    # Suppose there exist two unique solutions to the puzzle, `n` and `m`

    Soluton `n`:

    An, Bn
    Cn, Dn

    Solution `m`:

    Am, Bm
    Cm, Dm

    # Case 1 (Theorem: `An != Am`)

    Suppose `An = Am`

    Since `W = An*Bn = Am*Bm`, it follows that `Bn = Bm`, and so `Cn = Cm` and `Dn = Dm`, and so the solutions are NOT unique. (This would imply later a common factor of `1`).

    Therefore, `An != Am`, and so either `An < Am` or `An > Am`.

    # Case 2 (& 3)

    Suppose `An < Am` (If reversed, follows similarly.)

    Since `W = An*Bn = Am*Bm`, there must be some number `Fnm > 1` such that `Am = An*Fnm`. It follows that Bn = Bm*Fnm, and therefore `Bm < Bn`.

    Similarly, this `Fnm` factor passes between the `C` and `D` elements of the `n` & `m` solutions.

    The two solutions can be written in relation to each other and this common factor thusly:

    Soluton `n`:

    An, Bm*Fnm
    Cm*Fnm, Dn

    Soluton `m`:

    An*Fnm, Bm
    Cm, Dn*Fnm

    The Solution Product written in these terms is

    SP = An*Bm*Cm*Dn*Fnm^2

    # Conclusion

    For a prime factorization of `SP`, there will exist at least 1 prime with exponent 2 or greater. The product of these primes are potential values of `Fnm`.

    In case 1, `Fnm = 1`, therefore the two solutions were not unique. If a puzzle is defined to have the property that it has only one solution, then no common factor exists for either of the solution's diagonal pairs since the only allowable `Fnm` can be `1`.