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Posers and Puzzles

Posers and Puzzles

  1. 04 May '09 14:14 / 3 edits
    Is there a number when squared and when cubed all figures from 0 to 9 are represented.

    I give you a solution that is faulty:
    42^2 = 86420 and 42^3 = 97531. In these two answers combined every figure (0,1,2,3,4,5,6,7,8,9) is represented, even if they're not in the correct order. So one solution is 42.
    No, it is not, this is just an example. 42 is *not* a solution.

    Is there a solution? Are there many solutions? How did you do to find it out?
  2. Standard member forkedknight
    Defend the Universe
    04 May '09 20:24 / 1 edit
    69 is a solution. I found it with a script.

    There are many solutions if you allow extra characters, not sure if there are many if you don't.

    *edit* 69 is a unique solution if you allow only 10 characters total in the combined result (squared and cubed)
  3. Standard member forkedknight
    Defend the Universe
    04 May '09 20:30
    I figured this was an easy problem to solve with a script instead of mathematicaly since there are relatively few number that could even possibly provide a solution.

    n^2 must be 4 digits
    n^3 must be 6 digits

    That leaves possible solutions from 47 to 99 ... only 53 potential solutions.

    This is, of course, assuming that you don't allow repeated characters.
  4. 05 May '09 15:59
    forkedknight, you are quite right.

    If I change the question a little:
    Is there a number when qubed and when quadded all figures from 0 to 9 are represented. (n^3 and n^4)
    Again 42 is not a solution

    What about n^4 and n^5?
  5. Standard member forkedknight
    Defend the Universe
    05 May '09 16:26
    For n^3 and n^4, there are now only 4 possible solutions that have 10 digits: 18 through 21.

    18 happens to be a solution.

    For n^4 and n^5, there are no numbers that have a result of 10 digits.
    9^4 and 9^5 is 9 digits.
    10^4 and 10^5 is 11 digits.

    In fact, there are no solutions where the results of n^x and n^(x+1) uniquely represents all 10 digits beyond n^3 and n^4. This is simple to test exhaustively as 2^15 and 2^16 are 5 digits each, and anything beyond that will have more characters in the result.
  6. Standard member forkedknight
    Defend the Universe
    05 May '09 16:31
    I would have to say that this is a very interesting problem at first glance, but there are too many constraints to make it interesting mathematically because an exhaustive search is both faster and easier to do than using a mathematical approach.
  7. 06 May '09 01:28
    Originally posted by FabianFnas
    Is there a number when squared and when cubed all figures from 0 to 9 are represented.

    I give you a solution that is faulty:
    42^2 = 86420 and 42^3 = 97531. In these two answers combined every figure (0,1,2,3,4,5,6,7,8,9) is represented, even if they're not in the correct order. So one solution is 42.
    [b]No, it is not
    , this is just an example. ...[text shortened]... solution.

    Is there a solution? Are there many solutions? How did you do to find it out?[/b]
    not to be a troll, but i have to disagree with the way you posed the problem. more accurately, i think you are asking us to "prove that there are finitely many integers n such that n^2 and n^3 exhaust the digits, brute force methods not allowed." then, you ask if we can find one.

    however, as posed, your question is that of an existence proof: "is there an n such that n^2 and n^3 exhaust the integers?" and as such, 42 is a fine proof because an example proves that one such number exists.

    not trying to be a jerk, just kinda irked me i guess, that you said 42 is not a solution to your posed question... though, in context i see that you are asking for a more elegant and inclusive proof that accounts for all possible solutions.
  8. Subscriber joe shmo On Vacation
    Strange Egg
    06 May '09 01:51
    but 42 wasnt actually a solution...he just made up the numbers

    42^2 = 1764
    42^3 = 74088
  9. 06 May '09 04:38
    Originally posted by Aetherael
    not to be a troll, but i have to disagree with the way you posed the problem. more accurately, i think you are asking us to "prove that there are finitely many integers n such that n^2 and n^3 exhaust the digits, brute force methods not allowed." then, you ask if we can find one.

    however, as posed, your question is that of an existence proof: "is ther ...[text shortened]... are asking for a more elegant and inclusive proof that accounts for all possible solutions.
    No no, you're neither a troll,nor jerk. I see you quite friendly.

    The problem with a problem (?) is to formulate it correctly. A good formulation of a problem includes to define it properly. You gave me a good formulation. I chosed to formulate it as good as I could (despite my lack of English skills) and show an example how it could be done without any other help to the problem solver. So a faulty calculation of a faulty solution (42) gave a hint how I meant.

    I've seen many good problems ruined by poorly formulation, with side solution, pseudo solutions, and faulty solutions due to poorly formulated problems.

    So the problem #2 can be expressed: "prove that there are finitely many integers n such that n^3 and n^4 exhaust the digits, brute force methods not allowed."

    Btw - What's wrong with brute force? Ah, the interesting analysis is lost. For example: Every number ending with a zero gives a double zero when squared and therfore must be excluded. Are there more numbers that can be excluded?

    This problem is however a bagatelle. No deeper insights are needed. Excell can be a good tool.
  10. Standard member Palynka
    Upward Spiral
    06 May '09 10:34 / 1 edit
    Originally posted by FabianFnas
    No no, you're neither a troll,nor jerk. I see you quite friendly.

    The problem with a problem (?) is to formulate it correctly. A good formulation of a problem includes to define it properly. You gave me a good formulation. I chosed to formulate it as good as I could (despite my lack of English skills) and show an example how it could be done without an s problem is however a bagatelle. No deeper insights are needed. Excell can be a good tool.
    Numbers ending in 1 (or 5) will also repeat 1 (or 5) in the ^2 and ^3.
  11. 07 May '09 06:42
    Originally posted by FabianFnas
    No no, you're neither a troll,nor jerk. I see you quite friendly.

    The problem with a problem (?) is to formulate it correctly. A good formulation of a problem includes to define it properly. You gave me a good formulation. I chosed to formulate it as good as I could (despite my lack of English skills) and show an example how it could be done without an ...[text shortened]... s problem is however a bagatelle. No deeper insights are needed. Excell can be a good tool.
    i totally misunderstood what you were saying, mostly because i did not actually calculate 42^2 or 42^3 haha ... your formulation is just fine! i thought you were stipulating that just showing a number that fits the requirements is not enough, but that a "solution" needed to be presented showing why there are no more possible alternatives. sorry about that!
  12. Standard member uzless
    The So Fist
    15 May '09 18:52 / 1 edit
    fyi, in the hitchikers guide to the galaxy the answer 42 was misinterpreted by the audience after the computer gave 42 as the answer to the meaning of life.

    Keep in mind the author was british.

    The computer really said, "For tea, two"

    In other words, Tea for two was what life is all about. A typical british viewpoint on life and very clever by the author.


    Now you know.
  13. Standard member Palynka
    Upward Spiral
    15 May '09 19:06
    Originally posted by uzless
    fyi, in the hitchikers guide to the galaxy the answer 42 was misinterpreted by the audience after the computer gave 42 as the answer to the meaning of life.

    Keep in mind the author was british.

    The computer really said, "For tea, two"

    In other words, Tea for two was what life is all about. A typical british viewpoint on life and very clever by the author.


    Now you know.
    http://news.bbc.co.uk/1/hi/magazine/7287255.stm
  14. Standard member uzless
    The So Fist
    15 May '09 21:20
    Originally posted by Palynka
    http://news.bbc.co.uk/1/hi/magazine/7287255.stm
    I like tea for two best
  15. Standard member forkedknight
    Defend the Universe
    16 May '09 19:00
    Originally posted by uzless
    I like tea for two best
    The question provided by his series of books is something like, "what do you get when you multiply 6 by 9?". I always though this was hilarious because it would mean everything we think we know in math or science is wrong.