Please turn on javascript in your browser to play chess.
Posers and Puzzles

Posers and Puzzles

  1. Standard member royalchicken
    CHAOS GHOST!!!
    28 Dec '03 06:13
    This isn't mine. Suppose a positive integer is expressed in decimal notation. It is 'good' if the first n digits, read left to right, form a number divisible by n, for all n from 1 up to the total number of digits in the number. For example, 1236 is good since 12 is divisible by 2, 123 is divisible by three, and 1236 is divisible by 4. The puzzle is to find a number with each of the digits 1,2,3,4,5,6,7,8,9 used once, that is good.

  2. 28 Dec '03 10:10
    Originally posted by royalchicken
    This isn't mine. Suppose a positive integer is expressed in decimal notation. It is 'good' if the first n digits, read left to right, form a number divisible by n, for all n from 1 up to the total number of digits in the number. For example, 1236 is good since 12 is divisible by 2, 123 is divisible by three, and 1236 is divisible by 4. The puzzle is to find a number with each of the digits 1,2,3,4,5,6,7,8,9 used once, that is good.

    Is it allowed to use the digit zero? If not, does that mean you have to find an integer that consists of 9 different digits?
  3. Standard member TheMaster37
    Kupikupopo!
    28 Dec '03 11:19
    Originally posted by royalchicken
    This isn't mine. Suppose a positive integer is expressed in decimal notation. It is 'good' if the first n digits, read left to right, form a number divisible by n, for all n from 1 up to the total number of digits in the number. For example, 1236 is good since 12 is divisible by 2, 123 is divisible by three, and 1236 is divisible by 4. The puzzle is to find a number with each of the digits 1,2,3,4,5,6,7,8,9 used once, that is good.

    123456789 in any order form a multiple of 9. and the first number is always a multiple of 1.

    the second number has to be even, the third and fourth together must be a multiple of 4, the sixth, seventh and eighth together must be a multiple of 8. The sum of the first three digits must be a multiple of 3. And the fifth digit has to be a 5. the first six must be a multiple of 6, and thus the sum must be a multiple of 3. thus the 4th, 5th and 6th must be a multiple of three.

    In formulas;

    abcdefghi
    5|abcde => e=5
    2|ab => 2|b
    4|abcd => 4|cd
    8|abcdefgh => 8|fgh
    3|abc => 3|a+b+c
    6|abcdef => 3|d+e+f
    => 2|f
    7|abcdefg
  4. 28 Dec '03 11:39
    Originally posted by TheMaster37
    123456789 in any order form a multiple of 9. and the first number is always a multiple of 1.

    the second number has to be even, the third and fourth together must be a multiple of 4, the sixth, seventh and eighth together must be a multiple of 8. The sum of the first three digits must be a multiple of 3. And the fifth digit has to be a 5. the first six ...[text shortened]... 8|fgh
    3|abc => 3|a+b+c
    6|abcdef => 3|d+e+f
    => 2|f
    7|abcdefg
    With 7 digits I get 2 possibilities when I apply your formulas to it

    1296547 and 1472583.

    If this is right there is no solution. If I am wrong (more likely) I am curious.
  5. Standard member royalchicken
    CHAOS GHOST!!!
    28 Dec '03 16:42 / 1 edit
    There is a solution.
  6. Donation Acolyte
    Now With Added BA
    29 Dec '03 00:52 / 1 edit
    Anyone know a quick check for divisibility by 7?

    Edit: never mind, I used a calculator for that bit. I agree that a solution exists, namely 381654729. I believe it to be unique, but I might well be wrong on that, as I may have made a mistake in my ad hoc scribblings.
  7. Standard member royalchicken
    CHAOS GHOST!!!
    29 Dec '03 01:00 / 1 edit
    Take the number in question. Divide it by 7. If the fractional part is 0, you've lucked out .

    Seriously, try this. Take the last digit, and multiply it by two. Then subtract it from the number formed by truncating the last digit from the given number. Repeat until it is small enough to divide it. For example:

    1234 -> 123 - 8 = 115, 11 - 10 = 1, 1234 is not a multiple of seven.

    I think it works.

  8. Standard member StarValleyWy
    BentnevolentDictater
    29 Dec '03 01:10
    Originally posted by royalchicken
    Take the number in question. Divide it by 7. If the fractional part is 0, you've lucked out .
    a very "mod" answer!
  9. Standard member royalchicken
    CHAOS GHOST!!!
    29 Dec '03 01:15
    Hmmm...a Red Hot Pun...
  10. 01 Jan '04 15:59
    Originally posted by royalchicken
    There is a solution.

    And that is 381 654 729.

    An 'easy' way to find the solution for ABC DEF GHI is to start with the middle part.
    E = 5
    D and F must be even
    D+ E+ F is a multiple of 3

    This leaves 4 possibilities (258, 456, 654, 852)

    B, D, F, H are all even (2, 4, 6, 8)
    ? A, C, E, G and I are all odd
    CD is a multiple of 4
    And C is uneven, but not 5
    This leaves for CD 4 possibilities (32, 72, 16, and 96)

    This leaves 4 possibilities for CDEF
    1654, 3258, 7258 and 9654

    A + B + C is a multiple of 3
    A and C are odd (not 5)
    B is even
    This leaves 20 possibilities for ABC (321, 381, 741, 921, 981, 123,183, 723, 783, 963, 147, 327, 387, 927, 987, 129, 189, 369, 729, and 789),

    Combined this leaves 10 Possibilities for ABCDEF
    (321654, 381654, 921654, 981654, 963258, 147256, 129654, 189654, 729654 and 789654)

    Now you can see what is needed for G by dividing ABCDEFG by 7.
    That gives 6 possibilities (3816547, 9216543, 1472569, 1296547, 7296541 and 7296548)

    FGH must be divisible by 8.
    That leaves 38165472

    And the last number (I) must be a 9.

    Happy New Year
    Fjord
  11. Donation Acolyte
    Now With Added BA
    01 Jan '04 22:39
    Originally posted by fjord

    And that is 381 654 729.

    An 'easy' way to find the solution for ABC DEF GHI is to start with the middle part.
    E = 5
    D and F must be even
    D+ E+ F is a multiple of 3

    This leaves 4 possibilities (258, 456, 654, 852)

    B, D, F, H are all even (2, 4, 6, 8)
    ? A, C, E, G and I are all odd
    CD is a multiple of 4
    And C is uneven, but not 5
    This leave ...[text shortened]... le by 8.
    That leaves 38165472

    And the last number (I) must be a 9.

    Happy New Year
    Fjord
    I got there first
  12. 02 Jan '04 03:16
    Originally posted by Acolyte
    I got there first
    Stupid me I had missed that part of your post.
    All the honor to you
    Anyway by doing it again we have a 'scientific proof' that ihere is one and only one possibility

    Just now I threw the number in Google and found out that the number is rather famous.
    And I found there another puzzle, I thought was nice mentioning.

    What is the longest positive integer that is polydivisible (as above) and that is also a palindrome?

    E.g. 50405 is a palindrome and polydivisble.

    All digits (zero, 1,2,3,4,5,6,7,8,9) are permitted as many times as you want.
    To give you a hint: the solution exists of 11 digits

    Greetings, Fjord
  13. 02 Jan '04 16:09
    Seriously, try this. Take the last digit, and multiply it by two. Then subtract it from the number formed by truncating the last digit from the given number. Repeat until it is small enough to divide it. For example:

    1234 -> 123 - 8 = 115, 11 - 10 = 1, 1234 is not a multiple of seven.

    I think it works.

    what if the number ends with a zero?...ie., 1300?
  14. 02 Jan '04 17:04 / 1 edit
    Originally posted by kaushpaul
    what if the number ends with a zero?...ie., 1300?
    It still works. 1300, for example:

    130-(2*0)=130
    13-(2*0)=13
    1-(2*3)=-5, which is not divisible by seven, therefore 1300 is not divisible by seven

    Or, for 980:

    98-(2*0)=98
    9-(2*8)=-7, which is divisible by seven, therefore 980 is divisible by seven

    Cool trick!
  15. Standard member royalchicken
    CHAOS GHOST!!!
    02 Jan '04 20:54
    goldfish1, can you work out the mathematical basis of the trick? (Actually, you most likely can, but will you...?)