For divisibility by 13, take the last number, multiply by 4 and add to the rest.

For divisibility by 17, take the last number, multiply by 5 and subtract from the rest.

For divisibility by 19, take the last number, multiply by 2 and add to the rest.

In fact, you can adapt the method to check for divisibility by any prime number k.

There is a mathematical algorithm that proves this works in all cases. However this rule is not actually all that impressive as it appears at first glance! The number of operations (comparisons/subtractions/multiplications) you need to do is equivalent to just long-dividing the number by 7.

Consider: each operation of the rule removes one digit from the end. But you could just as easily apply the rule like "If the first digit is >=7, subtract 7 from it. Else, subtract the biggest multiple of 7 that will fit from the first two digits." To skip multiplying, you can use the following jump table: if the first digit is 6, subtract 54 from the first 2 digits, if 5 subtract 49, if 4: 35, if 3: 28, if 2: 14, if 1: 07. That will also remove one digit from the front! But now you are just doing long division.

https://math.stackexchange.com/questions/44459/proof-for-divisibility-by-7

It goes like this

1. create 2 distinct numbers by isolating the last digit from the other. For example, 154 becomes 15 and 4.

2. double the number derived from the last digit. So, the four becomes 8.

3. subtract from the number derived from the preceeding digits. 15 - 8.

4. the resulting number is 7. Seven is divisible by 7, so we know 154 is divisible by 7.

Yeah you got it wrong, it's

1 - 4*2 = -7

2 - 1*2 = 0

2 - 8*2 = -14