A more general divisibility rule, which applies to every power of two or five is that a number is divisible by 2^n if its last n digits are divisible by 2^n, and it's divisible by 5^n if its last n digits are divisible by 5^n. This follows from the way our base ten number system is constructed.
You are right! That's essentially how it is I just wanted to show you what goes on in the background. More rule proofs for divisibility by 3, 5, 8, 11 are coming soon!
Using the number from your thumbnail: 55557773336, there is a very intuitive proof. The last number 6 is divisible by 2, therefore the entire number is divisible by 2. Why? If I take away 6 from the original number I have: 55557773330 This is just a bunch of 10s (in fact it is exactly 5555777333 10s.) If 10 is divisible by 2 then every one of those 5555777333 10s can be divided by 2. The only thing remaining is the last digit, 6, which is divisible by 2. Now you may say it's not a real proof, but it is. Because I can theoretically layout 10 cases, for digits 0-9, and demonstrate whether or not these numbers can be split into two equal groups. Once those have been done manually, you can apply the rule above to solve for any given number.
