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Prove the following (how to tell if a number is divisible by 3)

  1. Aug 13, 2011 #1
    This thread has 2 parts.

    The first part consists of a problem I'm going to ask brains more intelligent than mine to help me solve.

    The second part consists of my wanting to know whether a person who has a BA in math who is this helpless when it comes to proofs should bother with grad school.

    The first part.

    Prove that if

    (x1 + x2 + ... + xn) % 3 = 0

    where xi is an integer such that 0 <= xi <= 9

    then the number whose digits equal those xi's, in that particular order, is also divisible by 3.

    This relates to that shorthand way most of us learned in school to tell if a number is divisible by 3. For example, how do we tell if the number 23520 is divisible by 3? We add up its digits and if the sum is divisible by 3, then so is the number. In this case 2 + 3 + 5 + 2 + 0 = 12, which means that 23520 is also divisible by 3.

    And now back to the second part of this post: given that I have a BA in math, and given that I spent almost 2 hours while I was driving on the highway trying to mentally solve this problem and couldn't do it, does it mean that I am not grad school material? Or am I trying to solve a problem that the average BA in math probably wouldn't be able to solve?
    Last edited: Aug 13, 2011
  2. jcsd
  3. Aug 14, 2011 #2
    EDIT: Sorry this doesn't answer your question at all!

    If X1%3=0 and X2%3=0 then (X1*10+X2)%3=0, i.e. (X1X2)%3=0 (these are separate digits in the last one, not a product)

    This idea can be extended to as many digits as you like. And to question number 2. Being hopeless at proofs means just means you got to put in more work. If you are willing to work hard then grad school will be easy. Oh yeah, and concentrate on the road when your driving!
  4. Aug 14, 2011 #3
    Attempt number 2:

    if (X1*10+X2)%3=0
    then (X1*9+X1+X2)%3=0
    then (X1+X2)%3=0
  5. Aug 14, 2011 #4
    Please allow me to rearrange your expression:
    is divisible by 3, then the number [itex]x=x_n x_{n-1}...x_3x_2x_1[/itex] is divisible by 3.

    If we add anything that is a multiple of 3 to [itex]x_d[/itex], it will still be divisible by 3. (property of the modular congruence, if you want to be technical). We can add [itex]9x_2[/itex], [itex]99x_3[/itex],[itex]999x_4[/itex] etc. to [itex]x_d[/itex] and get x. Since all of the things we added are divisible by 3, we have [itex]x\equiv 0 \mod 3[/itex].
  6. Aug 14, 2011 #5
    Although the previous two posts already pretty much answered it, I'm going to say it again, but just very concise (cause that's how I personally like it ;))

    [itex]x = \sum_{k=0}^{n} x_k 10^k \textrm{ is the number with those digits, and now } (x \mod 3) = \left( \sum_{k=0}^{n} (x_k\mod 3) (10 \mod 3)^k \right) = \left( \sum_{k=0}^{n} x_k \mod 3 \right) = 0[/itex]

    As for your second question: I'd say one problem is far from conclusive, try more :)
  7. Aug 15, 2011 #6
    The first time you saw an example finding the limit of a nontrivial equation at a point it was probably close to incomprehensible. Likewise the first time you saw an example of finding a derivative, an integral, a differential equation, etc it was probably about the same. The second or third or fourth time it was closer to just being a trick. The tenth or twentieth it was a method. I suggest that proofs may be the same.

    Consider that you are easily going to spend $25,000 on an in-state public grad school. You almost certainly do not want to get into that and then discover you are not going to make it. So I propose you spend perhaps five percent of that to get a much better idea whether you will succeed.

    I propose that you find a *qualified* tutor who is confident that in say 40-60 hours of their time, and *at least* several times that much of your time and hard work and concentration trying to learn a new and very foreign skill, that you will either become proficient in writing very good proofs or you will be much more certain that this is not the path for you.

    That might seem like a substantial investment just to answer a question, but consider how much even the first semester or two of grad school will cost you to find out the same answer and with a much higher personal cost if the answer turns out to be no.

    If most of the way through that tutoring process the answer appears to be more likely yes then you see if a local school has a very good prof teaching the first year of undergraduate analysis, or whatever they use for the first year of introduction to proof, get the textbook, explain to the prof you are looking at their grad school and politely ask if you might sit in the back, not take up any of their time, and work twice as hard as anyone else in class to see if you have what it will take.
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