wow !!!!

Discussion in 'Off-Topic Chat' started by little_miss_marcroft, Nov 11, 2004.

  1. theMouthPiece Related Searches

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  2. lynchie

    lynchie Active Member

    and now the maths geek in me wants to bust out and work out how that works....
     
  3. IckleSop

    IckleSop Active Member

    Hee hee That is kinda werid
     
  4. six pints

    six pints Active Member


    me 2, if i didnt have lecture in 10 mins i would!
     
  5. Griffin

    Griffin Active Member

    Didnt work when I typed in 666.

    Maybe its me? but does it work when all the numbers are the same?

    Or am I just a bit dumb?

    (I Pity the fool) - he said in his best Mr T voice!!
     
  6. Vickitorious

    Vickitorious Active Member

    hmmmmmmm :-?
     
  7. aimee_euph

    aimee_euph Member

    thart's not weird....it didn't work?
     
  8. drummergurl

    drummergurl Active Member

    thats well freaaaaaaaaaaaaaaaaaaaaky
     
  9. T-Horn

    T-Horn Member

    never worked!
     
  10. Naomi McFadyen

    Naomi McFadyen New Member

    I did it three times.... got me every time! :lol:

    Clever...
     
  11. theMouthPiece Related Searches

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  12. T-Horn

    T-Horn Member

    are you that bored?
     
  13. Euphliz

    Euphliz New Member

    hum.....................ah...................................yes.............no...............weird..................
     
  14. Mrs Womble

    Mrs Womble Member

    It's just been around the whole of my office!! very good :)
     
  15. lynchie

    lynchie Active Member

    I wouldn't trust your subtracting skills...
     
  16. rightnowmusic

    rightnowmusic Member

    Think it'll be something to do with multiples of "9"..... ish... :D
     
  17. T-Horn

    T-Horn Member

    leave me alone, it never worked. my subtracting was fine!!!!!!
     
  18. ComposerAndy

    ComposerAndy Member

    Yeah, something with multiples of nine. After taking one number from the other you can always add the digits together and get some multiple of nine, and so the number that's circled is whatever you need to add to the other two to make the next highest multiple of nine (55, you circle 8, 44, you circle 1 etc.) It always works out like this and so yes T-horn's subtracting skills suck
     
  19. six pints

    six pints Active Member

    someones feeling clever this morning!
     
  20. I know how it works!! *pulls her tongue out and does that "ner ner ni ner ner" thing*

    Go on then... I'll tell you....

    When you jumble up any large number and subtract the smaller one from the biggest one then then answer is always a number where the individual digits add up to a multiple of 9.

    (eg 3168 gives 3+1+6+8 = 18,
    or 2799 gives 2+7+9+9 = 27).

    If you select any three digits and add them together, then subtract the total from the next biggest multiple of nine you are left with the value of the remaining digit.
    eg if you select 3, 6 and 8, add them together = 17, subtract from the next multiple of 9 which is 18 then the answer is 18-17=1!

    So there!!

    *sniggers*.....
     
  21. MoominDave

    MoominDave Well-Known Member

    Almost a complete answer, but you didn't say why it's always a multiple of 9. And a true BOC abhors nothing like an incomplete BOCcy answer...

    If you take an n-digit number, A, with digits a{1}, a{2}, a{3}, ..., a{n} then you can write it as
    10^(n-1)*a{1}+10^(n-2)*a{2}+...+10*a{n-1}+a{n}

    If you muddle up the digits, you get a new n-digit number, B, with digits b{1}, b{2}, ..., b{n}, which can be written as
    10^(n-1)*b{1}+10^(n-2)*b{2}+...+10*b{n-1}+b{n}

    However, because the bs are the same as the as, but in some other order, we can also write B as
    10^m{1}*a{1}+10^m{2}*a{2}+...+10^m{n-1}*a{n-1}+10^m{n}*a{n}, where all of the ms are the integer powers from 1 to n-1, jumbled up to match the jumbling performed in changing A to B.

    Now, if we look at A-B (it doesn't matter which is larger - if something divides a number exactly, it will also divide minus that number exactly), then we see that we have the number
    (10^(n-1)-10^m{1})*a{1}+(10^(n-2)-10^m{2})*a{2}+...+(10^1-10^m{n-1})*a{n-1}+(10^0-10^m{n})*a{n}

    If we look at the ith term of this, we see that it is of the form
    (10^(n-i)-10^m{i})*a{i}.
    Just looking at the factor in brackets, we see that we have three cases: i) When n-i > m{i}, ii) When n-i < m{i} or iii) When n-i = m{i}.

    In case (i), the value of this can be written as 10^(n-i-m{i})*(10^m{i}-1). Looking at the bracketed bit in this, it is one less than a power of 10, i.e. a number of the form 9999...9, which evidently divides by 9 (giving the answer 1111...1). So the ith term of our expression divides by 9.

    In case (ii), it is similar, except that we write the factor as 10^(m{i}-(n-i))*(10^(n-i)-1).

    In case (iii), the factor is zero (10^x-10^x).

    So, every term is divisible by 9, and hence the whole difference of the two jumbled numbers is divisible by 9, for any number, of any number of digits, not just 3 or 4.

    Ahthankyewverymuch...
     

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