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Math Help - Very Hard Math Puzzle

  1. #1
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    Very Hard Math Puzzle

    So you have 6 numbers, ranging from 1-24. Five of the six numbers must equal the 6th number by using any of the 4 operations (+, -, *, /) and using each number only once.

    So for example:

    3 7 12 19 21 | 5

    So you need to get the number 5 by moving around and using the operations on those five first numbers.


    Now, find an example where this is impossible!

    EDIT: I think I found one:

    5 11 17 19 23 -> 1

    See if you can find away this would work, cause I cannot!

    Also, how would one go about proving it is a valid counter example?
    Last edited by stones44; January 11th 2010 at 05:02 PM.
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    Suggestion

    Shouldn't this go under the Math Challenge Problems section?
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  3. #3
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    Quote Originally Posted by stones44 View Post
    so you have 6 numbers, ranging from 1-24. Five of the six numbers must equal the 6th number by using any of the 4 operations (+, -, *, /) and using each number only once.

    So for example:

    3 7 12 19 21 | 5

    so you need to get the number 5 by moving around and using the operations on those five first numbers.


    Now, find an example where this is impossible!

    Spoiler:
    1,5,9,17,23 | 24
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    1,5,9,17,23 | 24
    Surely (23-17)*(9-5)/1=24... you need to find something else, 24 has too many factors
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    consider 1 6 11 16 21 | 18

    I've thought it over many times and think it works well as a counter example
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    Quote Originally Posted by sym0110 View Post
    Surely (23-17)*(9-5)/1=24... you need to find something else, 24 has too many factors
    For some reason in my head I was thinking you can only add or subtract a 1, because multiplication or division by 1 were pointless Nevermind.


    I don't think 24 having too many factors is a fair reason to rule it out though, if you pose the same problem but with 4 numbers instead of 5, then there are plenty of combinations of 4 numbers that can never be made to equal 24.
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    Quote Originally Posted by sym0110 View Post
    consider 1 6 11 16 21 | 18

    I've thought it over many times and think it works well as a counter example
    A counterexample to what?

    If you mean a solution to the problem, you missed:

    21 - (16 - 1) / (11 - 6) = 18
    6 * ((21 + 11) / 16 + 1) = 18
    11 * (6 - 1) - 21 - 16 = 18
    21 - (16 + 6) / 11 - 1 = 18
    6 * (21 - 16) - 11 - 1 = 18
    16 + (21 - 11) / (6 - 1) = 18
    Last edited by pomp; January 9th 2010 at 03:09 PM.
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    EDIT: I think I found one:

    5 11 17 19 23 -> 1

    See if you can find away this would work, cause I cannot!

    Also, how would one go about proving it is a valid counter example?
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    Quote Originally Posted by stones44 View Post
    edit: I think i found one:

    5 11 17 19 23 -> 1

    see if you can find away this would work, cause i cannot!

    also, how would one go about proving it is a valid counter example?
    (17 + (19 + 11) / 5) / 23 = 1
    17 / ((19 - 11) * 5 - 23) = 1
    ((23 - 17) * 5 - 11) / 19 = 1
    (11 + (23 + 17) / 5) / 19 = 1
    11 / ((23 - 17) * 5 - 19) = 1
    (19 - 11) * 5 / (23 + 17) = 1
    (23 - 17) * 5 / (19 + 11) = 1
    (19 + 11) / (23 - 17) / 5 = 1
    ((19 - 11) * 5 - 23) / 17 = 1
    (23 - (19 + 11) / 5) / 17 = 1
    23 / ((19 - 11) * 5 - 17) = 1
    11 / (19 - (23 + 17) / 5) = 1
    19 / (11 + (23 + 17) / 5) = 1
    (23 - 11) / (19 - 17) - 5 = 1
    17 / (23 - (19 + 11) / 5) = 1
    23 / (17 + (19 + 11) / 5) = 1
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    how are you coming up with all these so quickly? do you have some method or program?

    also how could one go about proving a counter example? (or proving there are none)
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    Quote Originally Posted by stones44 View Post
    how are you coming up with all these so quickly? do you have some method or program?

    also how could one go about proving a counter example? (or proving there are none)
    I have written a program that can check them.

    As for proving a counter example or proving there are none, that's rather difficult I think.

    The only way I can think of showing that a solution is true is by an exhaustive algorithm. If someone were to say they have a solution, I can verify it computationally using a program that exhausts all possible parsings of the numbers and symbols, and shows none of them equal the target number.

    Whether or not you see this as a proof is your call, you'd have to have confidence that my program really did check all possibilities.
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    Quote Originally Posted by pomp View Post
    I have written a program that can check them.

    As for proving a counter example or proving there are none, that's rather difficult I think.

    The only way I can think of showing that a solution is true is by an exhaustive algorithm. If someone were to say they have a solution, I can verify it computationally using a program that exhausts all possible parsings of the numbers and symbols, and shows none of them equal the target number.

    Whether or not you see this as a proof is your call, you'd have to have confidence that my program really did check all possibilities.
    True there aren't many ways of organizing the +-/* and brackets with 4 numbers, and there are only 24C5=42504 ways to exhaust all possibilities. It's probably only going to take a few hours to check all of them with a not so efficient program. But I don't think computer provided exhaustion is really a proof...
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    Quote Originally Posted by sym0110 View Post
    True there aren't many ways of organizing the +-/* and brackets with 4 numbers, and there are only 24C5=42504 ways to exhaust all possibilities. It's probably only going to take a few hours to check all of them with a not so efficient program. But I don't think computer provided exhaustion is really a proof...
    But we're using + - / * on 5 numbers, with a 6th as a target. There are so many combinations.

    Why wouldn't you see this as proof? Do you think the 4 colour theorem proof is not really a proof? It's not the most elegant of proofs, actually, it's probably the most inelegant proof I've ever heard of, but it's proof nonetheless.
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