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

  • Thread starter Ocerox
  • Start date
In summary, this conversation discusses a problem related to proving divisibility by 3 and whether a person with a BA in math should pursue grad school despite struggling with proofs. It is suggested that the person invest in tutoring to improve their proof-writing skills before committing to grad school.
  • #1
Ocerox
1
0
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:
Mathematics news on Phys.org
  • #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!
 
  • #3
Attempt number 2:

if (X1*10+X2)%3=0
then (X1*9+X1+X2)%3=0
then (X1+X2)%3=0
 
  • #4
Please allow me to rearrange your expression:
If
[tex]x_d=x_n+x_{n-1}+...+x_3+x_2+x_1[/tex]
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].
 
  • #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 :)
 
  • #6
Ocerox said:
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?

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.
 

What is the definition of divisibility by 3?

Divisibility by 3 means that a number can be evenly divided by 3 without any remainder.

What is the rule for determining if a number is divisible by 3?

The rule for determining if a number is divisible by 3 is to add up all the digits in the number. If the sum is divisible by 3, then the original number is also divisible by 3.

Can a number be divisible by 3 and not have a sum of digits divisible by 3?

No, a number cannot be divisible by 3 without having a sum of digits divisible by 3. If the sum of digits is not divisible by 3, then the original number is also not divisible by 3.

What are the exceptions to the rule of divisibility by 3?

The only exceptions to the rule of divisibility by 3 are when the sum of digits is 0 or 3. These numbers are automatically divisible by 3.

How is divisibility by 3 used in mathematics and science?

In mathematics and science, divisibility by 3 is used to simplify and solve equations, identify patterns in numbers, and solve problems related to fractions and decimals.

Similar threads

Replies
4
Views
224
Replies
7
Views
3K
Replies
2
Views
1K
  • General Math
Replies
11
Views
5K
  • General Math
Replies
7
Views
2K
  • General Math
Replies
13
Views
1K
Replies
4
Views
1K
Replies
4
Views
2K
Replies
8
Views
6K
Replies
2
Views
2K
Back
Top