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Posers and Puzzles

Posers and Puzzles

  1. 27 Apr '08 22:38
    This number has 9 digits, uses 1-9, and repeats no numbers. It is divisible by 9. If the most right digit were not there, the number would be divisible by 8. If this step were repeated on the next number, the remaining number should be divisible by 7. This goes on until the last number would have 1 digit and be divisible by 1.

    Please show how you got there, and your solution, and I will tell you if you are correct. It took me about 10 minutes to figure this out, so I will post the answer (if no one is right) in about 10 days.
  2. 27 Apr '08 23:16
    My start is that the number is of the pattern OEOE5EOEO where O is odd digit, and E an even, due to divisibility by even numbers and 5.

    Furthermore, 2 and 6 occupy the second and last even spot, as both of these are the only numbers which are divisible by 4 when they follow an odd number, which leaves 4 and 8 for the 1st and 3rd in some order.
  3. 27 Apr '08 23:22
    Originally posted by geepamoogle
    My start is that the number is of the pattern OEOE5EOEO where O is odd digit, and E an even, due to divisibility by even numbers and 5.

    Furthermore, 2 and 6 occupy the second and last even spot, as both of these are the only numbers which are divisible by 4 when they follow an odd number, which leaves 4 and 8 for the 1st and 3rd in some order.
    You got the OEOE5EOEO right, but I am not sure what you mean by the next part? Could you clarify?
  4. 27 Apr '08 23:26
    Originally posted by UzumakiAi
    This number has 9 digits, uses 1-9, and repeats no numbers. It is divisible by 9. If the most right digit were not there, the number would be divisible by 8. If this step were repeated on the next number, the remaining number should be divisible by 7. This goes on until the last number would have 1 digit and be divisible by 1.

    Please show how you got th ...[text shortened]... 10 minutes to figure this out, so I will post the answer (if no one is right) in about 10 days.
    381654729

    I don't know if this is a unique solution. My method was as follows:

    A) Observe that divisibility by 9 for the whole thing is inevitable - the digits sum to 45, a multiple of 9. Divisibility by 1 for the sole digit is also obviously inevitable.

    B) Positions 2, 4, 6 and 8 had to be evens, and the rest odd, fairly obviously.

    C) Position 5 had to be 5.

    D) Position 4 had to be either 2 or 6. The reason for this is that any odd digit followed by a 2 or 6 will be a multiple of 4, but an odd digit followed by a 4 or 8 won't. For divisibility by 4, only the last two digits are relevant.

    E) Similarly, but slightly less obviously, because position 6 is even, and hence position 6 multiplied by 100 is a multiple of 200 and hence of 8, it follows that positions 7 and 8 combined had to be a multiple of 8. Stripping out those with doubled up digits and zeros, that means only 16, 32, 72 or 96 were candidates, which makes position 8 either 2 or 6.

    F) Positions 1, 2 and 3 had to sum to three. Because the three-digit number was a multiple of three, the 6 digit number would be a multiple of three, multiplied by 1,000, plus three more digits, which meant that the number at the next three positions had to be a multiple of three, so digits 4,5 and 6 had to sum to three. By the same logic, so did the digits 7 through 9.

    G) That left me the following:

    Digits 4,5,6 either 654 or 258.
    Digits 7,8,9 either 321,327,723,729 or 963.
    The remaining even digit would go into position two. The remaining odd digits would go into positions 1 or 3.

    That narrows it down to at most 2x5x2 = 20 options.

    At that point I got lazy and built a spreadsheet to do trial or error and see which of these permutations divides by 7. 381654729 was the first one I found. It wouldn't take long to check the rest. I can't see how to narrow it down much further than that. I am not aware of any simple test for divisibility by 7.
  5. 27 Apr '08 23:29 / 1 edit
    There is a test for divisibility by seven.

    Remove the ones digit, and double it. Then find the difference between it and the rest of the number. If it is divisible by 7, then the original number is as well.

    For instance, 14. Taking the ones off and double leaves you 1 and 8, a difference of 7.

    112 splits into 11 and 2*2=4. 11-4 = 7, and 112 = 7 * 16.
  6. Subscriber joe shmo
    Strange Egg
    28 Apr '08 00:08
    Originally posted by UzumakiAi
    This number has 9 digits, uses 1-9, and repeats no numbers. It is divisible by 9. If the most right digit were not there, the number would be divisible by 8. If this step were repeated on the next number, the remaining number should be divisible by 7. This goes on until the last number would have 1 digit and be divisible by 1.

    Please show how you got th ...[text shortened]... 10 minutes to figure this out, so I will post the answer (if no one is right) in about 10 days.
    Are you intending to be arrogant? "It took me about 10 minutes to figure this out, so I will post the answer (if no one is right) in about 10 days.".... If your so in love with yourself the two of you should get a room..
  7. 28 Apr '08 00:10
    Originally posted by d36366
    381654729

    I don't know if this is a unique solution. My method was as follows:

    A) Observe that divisibility by 9 for the whole thing is inevitable - the digits sum to 45, a multiple of 9. Divisibility by 1 for the sole digit is also obviously inevitable.

    B) Positions 2, 4, 6 and 8 had to be evens, and the rest odd, fairly obviously.

    C) Position 5 ...[text shortened]... row it down much further than that. I am not aware of any simple test for divisibility by 7.
    Spreadsheet is interesting idea. 381654729 is a unique solution, and you got it! Good job...
  8. 28 Apr '08 04:26
    Aside from the 7-digit divisible-by-7 condition (the hardest to check), there are 6 possible numbers.

    147,258,963
    183,654,729
    381,654,729
    741,258,963
    789,654,321
    987,654,321

    Of these, 381,654,729 is the only one which meets the most difficult to check condition.
  9. 28 Apr '08 07:50
    Originally posted by geepamoogle
    Aside from the 7-digit divisible-by-7 condition (the hardest to check), there are 6 possible numbers.

    147,258,963
    183,654,729
    381,654,729
    741,258,963
    789,654,321
    987,654,321

    Of these, 381,654,729 is the only one which meets the most difficult to check condition.
    If you exclude the 7 condition, there are four more answers than that:

    981,654,327
    981,654,723
    189,654,327
    189,654,723

    I am pretty sure that is all of the options. See my post above which suggests 20 possibilities to check, but in fact ten of them have a repeated digit.

    And there is indeed only one solution once you include the 7 condition.