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Prove or Disprove

  1. Apr 28, 2004 #1
    PROBLEM: For any number in any base at least 2 digits long, if you add them together and solve it the answer will ALWAYS be 1 number less than the base itself.

    Why this problem works out?
    Does it apply to anything?
    What branch of Mathematics?
    Prove in General sense or disprove.

    HELP!!! :eek:
     
  2. jcsd
  3. Apr 29, 2004 #2

    matt grime

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    'solve'? what do you mean? post an example.
     
  4. May 3, 2004 #3
    I asked and solve it just means figure out if it works or not......
     
  5. May 3, 2004 #4

    matt grime

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    your question still doesn't make sense at all.

    take 11 (we won't say what base), what must we do to it now? what on earth do i add together? the only thing can i can think of is the digits in the number, so i add them and get 2, now in any other base than 3, that cannot be one less than the base, so what on earth are you talking about? at least attempt to indicate what the referents of all those undetermined pronouns are.
     
  6. May 3, 2004 #5
    I'm going to email my prof. and make sure I copied the question down right. i'll re-post it as soon as I hear from him. Thanks for your interest in helping.
     
  7. May 5, 2004 #6

    HallsofIvy

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    It is true that if you add all the "digits" in a number (in any base) and repeat until you are down to one digit, that digit will be the remainder when the original number is divide by the base minus 1. In base 10, it's called "casting out nines".
     
  8. May 5, 2004 #7
    I asked my teacher for an explanation and this is what he gave me.........

    Choose a number in any base that is two or more digits long.
    Add all the digits together and keep adding until you have 1 single digit.
    remember to add in the rules for the base you've chosen.
    Eg: Base 10
    743=7+4+3=14=1+4=5
    Take your original number (above it's 743) and subtract the "sum"
    743-5=738
    "Add" this "sum" together until you have 1 digit (as above)
    7+3+8=18=1+8=9
    This number is always one less than the base you're working in.

    Note: Your math MUST be consistant for the base you're working in!
     
  9. May 5, 2004 #8

    matt grime

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    well, now you've explained it it becomes reasonably obvious if you're happy with modulo arithmetic. as hallsofivy states the number you calculate in this operation is the reaminder modulo the base minus one, so subtracting it leaves a number that has zero remainder modulo the base minus one.
     
  10. May 6, 2004 #9

    Gokul43201

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    Your questions still appear, largely, unanswered... so let me fill in a little more :

    1. Why it works out can be proved using algebra and the representation of a number in a base, b as :
    sum [a(k) b^k], k=0,1,..n.

    It's easy to prove the part that hallsofivy stated. That comes from (b^m - 1) being divisible by (b-1); b being the base. The next step is to prove that the (sum of digits of a multiple of (b-1)) = a multiple of b-1. The only trick here is in noticing that some digits look like a(k) - a(k-1), which can be negative. So, you need to carry over a 'b' from the previous digit (to make it non-negative) while adding digits, and then everything will be okay.

    2. Does this apply to anything ? Yes, it applies to all numbers in Z. But if you're asking if it has any practical applications, I'm not sure I know any. Didn't Hardy once say that he felt secure that Number Theory would never be defiled by the common man, because it had no known applications ? RSA sure proved him wrong !

    3. I guess you might say this came under number theory.

    4. It can be proved (in the affirmative). For hints, see the answer to 1. If you really want to see the whole proof, I'll do it when I find an equation editor I can paste off of.
     
  11. May 6, 2004 #10

    matt grime

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    Equation editor? Why? Five minutes, and one pencil ought to do. The proof can be easily filled in from what has been written. Easily means after a day of thinking.
     
  12. May 6, 2004 #11

    Gokul43201

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    Actually, it's easier to prove using modular arithmetic, particularly, b^k==1 (mod b).
     
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