Okay, so before I begin (my first ever blog post!), a disclaimer. Firstly, I'm by no means a professional mathematician. I won't be using any fancy methods: really, this is my way of trying to hone my own mind, and explore the best processes for logic. But, if this does help anyone (or provide entertainment!) I am eternally grateful. I try to design these in a way that's accessible, explaining my thoughts step-by-step, as it were. Anyways, boring stuff over! Onwwards!

So, the Divisible by 3 rule. What is it? Well, if you read my blurb on the blog page you'd know, but perhaps if anyone ends up here by miracle from a search engine, I ought to explain. If you add the digits of a number, and they are divisible by 3, the entire number is too. Why? Well, that's what I'll be proving.
When I found this rule I wanted my own proof, so I made one. It's not as elegant as the Khan Academy one and such, and in fact it might be wrong. But it seems to work in my eyes, so let me know what you think!

So, the first thing I needed to consider, was how I could tie together the act of adding the digits and that of dividing the actual number.. Let's consider the latter point first. What actually happens when we divide a number? Well, if you use the Bus-stop division method, you go through each digit, from thousands, to hundreds, to tens, etcetera, and imagine that it's just a unit. Then, whatever the remainder is, we append to the next number, and continue. Of course, the usage of the remainder is just a way of counteracting that idea that we're dividing a unit, because we're not. Take, for example, 534/3, which is equal to 178. What we're doing when dividing this is: $$\frac{500}{3}+\frac{30}{3}+\frac{4}{3}$$ which is in itself equal to $$166.\dot{6}+10+1.\dot{3}$$ which adds up to $$177.\dot{9}$$ Because that goes on forever, it's effectively equal to $$178$$. This is where it gets complicated, because we leave the realm of common sense (where, in fact $$177.\dot{9}$$ is not equal to 178. Infinity is a tricky concept in maths, and that's what the recurrence symbol gives us. If, however, you put that equation (500/3+10...) into a calculator, it will confirm our summation. Okay, so we know how to divide by 3 in a way 10x more complicated then before. How does this help us? Well, okay. Let's consider what we're doing when adding up our digits and dividing them by 3. Because effectively we're doing the same thing, but we're dividing by a greater number to ensure we're treating it like units. So we end up with: $$\frac{500}{300}+\frac{30}{30}+\frac{4}{3}$$ Note how the denominator is being divided by 10 each time? What this really is, is: $$\frac{5}{3}+\frac{3}{3}+\frac{4}{3}$$ which adds to $$4$$, which as it's a whole number shows that the summation of the digits is also divisble by 3. Okay, wait a second. Let's confirm that in our brain quickly. $$5+3+4=9$$ which is obviously divisible by 3. But, we still need to know why. And why, more importantly, is this justa special case for 3? How do we know that this is not going to be hoodwinked, as it were, by another number?

Right. Let's take a step back. There's one key property of 3 that is vitally important for this to work. It's that, if a number is not divisible by 3, then it will always return a recurring decimal. If a number is not divisible by 3, it must be either 1 or 2 from the multiplier of 3 before it. In the case of 1, it returns a recurring 3, in the case of 2 it returns a recurring 6. And, as we know, these will sum to a whole number if added in the right way. So, let's go back to our adding of the digits. Looking at it now, I can actually simplify it so that they all have the same denominator: $$\frac{500}{500}+\frac{300}{300}+\frac{400}{300}$$ which gives it a look closer to what we have when we divide the number by 3 normally. Interesting... But what can I do now? This, in real life, was where I had my epiphany. I was expecting to have to do some pretty complex maths, but instead I just needed common sense. Let's take a look at the original dividing process again: $$\frac{500}{3}+\frac{30}{3}+\frac{4}{3}$$ And now, let's multiply our dividing process for the digits by 100 (humour me). Note that the result will go up to 400, but we don't care about that: we're trying to proof this for every multiple of 3. So, now (after simplifying the multiplication), we have $$\frac{500}{3}+\frac{300}{3}+\frac{400}{3}$$. Looking at the normal multiplication, these are really, really similar. In fact, all we've done is multiplied the 2nd and 3rd fraction by 10 and hundred respectively. Let's take a moment to consider what that means. The recurring decimal goes on for infinity, and because it's a single recurrence (e.g. not .676767, it's just .6666). So when we multiply those two numbers, what comes after the decimal point doesn't actually change at all. So (and here comes the anticlimax) if it's what's after the decimal point that counts to make a whole number, and what's after the decimal point is exactly the same as what it is when we're usually dividing, then if the normal dividing process returns an integer, so must adding the digits! And so, adding them and getting an integer after dividing by 3, proves all along that the entire number was too.

I hope you found that as anticlimatic as I did. But, it makes sense, which is what's great. The key thing in this proof is the properties of 3's recurring decimal quotients. We don't need to bother if there isn't a decimal, because then it'll add to a whole number anyway. I hope this made sense, if you read this through! And I hope this holds up if someone else reads this haha. If it doesn't, shoot me an email. But all the same, I hope you enjoyed!