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Posers and Puzzles

Posers and Puzzles

  1. Subscriber joe shmo On Vacation
    Strange Egg
    27 Sep '08 18:10 / 1 edit
    if you(ie.me) don't know number theory , it's not.

    here goes, is there an algebraic proof, so to speak, for this property?

    56 + 9 = 65

    I've figured out this much, but its not generalized.

    10a + b + 9 = 10b + a...........where b = (a+1)

    10a + (a+1) + 9

    10a + a + 10

    10(a+1) + a

    10b + a = 10b + a

    there is also the reverse where 10a + b -9 = 10b + a, if b = (a-1).

    but how is it proven for any number of consecutive digits, when the number added or subtracted = 9*10^(k-2), where = k the number of digits
  2. Standard member eldragonfly
    leperchaun messiah
    27 Sep '08 20:44
    You need to visit The Church of the Flying Spaghetti Monster today.
  3. Subscriber joe shmo On Vacation
    Strange Egg
    27 Sep '08 21:21
    Originally posted by eldragonfly
    You need to visit The Church of the Flying Spaghetti Monster today.
    and why should I do that?
  4. Standard member eldragonfly
    leperchaun messiah
    27 Sep '08 21:35 / 1 edit
    Originally posted by joe shmo
    and why should I do that?
    Bad joke. i don't understand what relationship you are trying to show, i can't see the connection that you are trying to show/imply.
  5. Subscriber joe shmo On Vacation
    Strange Egg
    27 Sep '08 21:39
    Originally posted by eldragonfly
    Bad joke. i don't understand what relationship you are trying to show, i can't see the connection that you are trying to show/imply.
    oh,...so my math is not clear.....

    if the digits are consecutive ....56, 45,34,67,89,23,ect....then adding 9 results in the reverse of the consecutive digits...so the first row plus nine = 65,54,43,76,98,32

    hope that clears it up
  6. Standard member wolfgang59
    Infidel
    29 Sep '08 11:40
    Originally posted by joe shmo
    oh,...so my math is not clear.....

    if the digits are consecutive ....56, 45,34,67,89,23,ect....then adding 9 results in the reverse of the consecutive digits...so the first row plus nine = 65,54,43,76,98,32

    hope that clears it up
    you are trying to prove that

    {10a + (a+1)} + 9 = {10(a+1) + a}

    since both sides of the equation are 11a + 1 its QED

  7. 01 Oct '08 02:27
    >>how is it proven for any number of consecutive digits, when the number added or subtracted = 9*10^(k-2), where = k the number of digits

    It does not hold in general, for k=3 for instance 123 + 90 = 213
  8. Subscriber joe shmo On Vacation
    Strange Egg
    01 Oct '08 02:41
    Originally posted by apawn
    >>how is it proven for any number of consecutive digits, when the number added or subtracted = 9*10^(k-2), where = k the number of digits

    It does not hold in general, for k=3 for instance 123 + 90 = 213
    ahh,..I guess if it did 123+90 would have to = 321

    ok, Thanks