Multiples of 3 are very neat. You've probably heard that you can add the digits of a number together, and if they result in a multiple of 3, then the number is a multiple of 3 which is very cool! But addition is too slow, and I've noticed few other patterns that can help you spot a multiple of 3 even faster which I've written below. It's a terribly nerdy ramble with some very rushed explanations, so read at your own risk. (Also, this isn't really groundbreaking or interesting, I was just terribly bored in middle school and liked playing with patterns in my head. Honestly this may be taught to some people, but it wasn't to me.)
I'm going to call the digits 1, 4, and 7 Group One (G1), 2, 5, and 8 Group Two (G2), and 3, 6, 9 and 0 as Group 3 (G3). You can use these groups to quickly identify if a number's a multiple of 3:
First, remove all the digits from G3:
Next, if a number from G1 is next to a number from G2, remove both of them:
You'll repeat the process above until you either have 0 digits, or digits all from the same group as I have above. Now, if you have 3 digits all from the same group, you can remove those as well, and keep doing as such until you only have 2, 1, or 0 digits. Given my number above, I can remove 885 and be left with 0 digits. If you have 0 digits left, congratulations! You have a multiple of 3. Otherwise, your number is not divisible by 3. Since I'm left with 0 digits, I know that 387880564978950 is divisible by 3.
This may seem a bit overwhelming at first, but with practice it becomes very easy to do it quickly in your head, and it's much faster than summing up all the digits or actually dividing by 3!
To explain how this works, the method is based on the idea that if the digits add up to a multiple of 3, then the number is divisible by 3. I started by removing all of G3 since any multiple of 3 plus another multiple of 3 will be another multiple of 3. (3x+3y=3(x+y)) Therefore, any digits from G3 will not make an impact on if the digit is a multiple of 3, and therefore we can ignore them.
The G1 digits can all be represented as n+1 where n is some multiple of 3 (0+1=1, 3+1=4, 6+1=7). The G2 digits can all be represented as m-1 where n is some multiple of 3 (3-1=2, 6-1=5, 9-1=8). Therefore, a digit from G1 plus a digit from G2 can be represented as n +1 +m -1 or n+m, and since n and m are both multiples of 3, their sum will also be a multiple of 3. Hence why you can remove a pair of G1 and G2.
Finally, if you have 3 digits from the same group they'll also result in a multiple of 3. 3 digits from G1 can be represented as n(1)+1+n(2)+1+n(3)+1, or n(1)+n(2)+n(3)+3 where n(x) is the multiple of 3 for digit x. As you can see, all these numbers are divisible by 3, and therefore the sum will be divisible by 3. When three of G2 is added up it results in m(1)+m(2)+m(3)-3 where m(x) is the multiple of 3 for digit x, and we can apply the same logic with G1 here.
There's more you can do here with multiples of 3 and these groups, but this post is long enough so I'll save it for later. Hopefully this makes sense and I didn't make some horrible mistake while writing this at midnight.
