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Number theory problem

  1. Mar 13, 2007 #1
    came across this in a book im reading, it doesnt give the answer though youll know once you get there that you're right anyways.

    1. The problem statement, all variables and given/known data

    In a parlour game, the 'magician' asks one of the participants to think of a three-digit number [tex]abc_{10}[/tex]. Then the magician asks the participant to add the five numbers [tex]acb_{10},/bac_{10},/bca_{10},/cab_{10},/cba_{10}[/tex], and reveal their sum. Supppose the sum was 3194. What was the [tex]abc_{10}[/tex] originally?

    2. Relevant equations

    none known

    3. The attempt at a solution

    right, Im not positive I'm doing the right thing, but I hope I am on the right track, just say where I need to go if I'm not, or correct me, or do whatever, thnx

    [tex]acb_{10} + bac_{10} + bca_{10} + cab_{10} + cba_{10} = 3194[/tex]

    expand that

    [tex]100a + 10c + b + 100b + 10a + c + 100b + 10c + a + 100c + 10a + b + 100c + 10b + a = 3194[/tex]

    [tex]122a + 212b + 221c = 3194[/tex]

    now, from here I'm kinda stuck. I have a hunch I have to use modular arithmetic somewhere (only just roughly learnt it so that why I kinda stuck on how to use it).

    If I was to guess I'd do something like

    [tex]a (mod/122) + b (mod/212) + c (mod/221) = 3194[/tex]

    but even with that guess id be stuck.

    I 'could' try do a lot of guesswork and go from there but I'm guessing there's a more elegant way. thnx

    hope you can help
     
  2. jcsd
  3. Mar 13, 2007 #2

    arildno

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    Note that if you add all six permutations together, you get 222(a+b+c) as your answer.
    Now, (a+b+c) is the cross-sum of each permutation, and the trick most likely consists in finding this cross-sum by aid of the known number.

    Thus, given that you have found this, multiply it by 222 and subtract the sum you were told to find the original number.

    Or something along these lines..
     
  4. Mar 13, 2007 #3
    soz, but what do you mean by 'cross-sum'?

    thnx
     
  5. Mar 13, 2007 #4

    Dick

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    He means 'cross sum'=a+b+c. So he has shown 222*(a+b+c)-3194=abc. On way to go is to reduce mod 222. So you get -86=abc(mod 222). Which gives you only 4 possibilities 136, 358, 580, 802. Then using your first equation you can pretty quickly figure 358 is the right one. But that seems like a laborious trick. It would be nicer if you could find a clever way to deduce a+b+c directly. Then you wouldn't have to guess.
     
  6. Mar 13, 2007 #5
    ooo rite, i see what you did there, everything bar the bit where you took

    222*(a+b+c)-3194=abc and reduced it mod 222 to -86=abc(mod 222)

    but i see how you attained the 4 possibles, so could you quickly explain the bit i just mentioned? thnx
     
  7. Mar 13, 2007 #6

    Dick

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    222*(a+b+c)=0 mod 222, right?
     
  8. Mar 14, 2007 #7
    o rite yeh, soz i see it now,

    thnx for all the help with it

    now i knew that i had to play with mod 222 I started from scrathc and did it all myself and got it right :D yey go me

    thnx
     
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