## Divisibility Rules

Cool Divisibility Rules

Special Thanks to: Patrick Burnett

You would probably learn the divisibility rules for 2, 3, 4, 5, 6, 8, and 9 in school. But what about the divisibility rule for 7? Well, the divisibility rule for 7 is quite simple, and quite interesting. All you have to do is take off the last digit of the number, multiply it by 2, and subtract that from the rest of the number. Here’s an example: Say you want to know if 469 is divisible by 7. If 469 is divisible by 7, then 46 – 2×9 must also be divisible by 7, and 46 – 18 = 28. Since 28 is divisible by 7, 469 is divisible by 7. That’s a quick way to check divisibility without having to do long division. Here’s another example: You want to know if 999999 is divisible by 7. If 999999 is divisible by 7, then 99999 – 2×9 = 99981 must be divisible by 7, and if 99981 is divisible by 7, then 9998 – 2×1 = 9996 must be divisible by 7, and if 9996 is divisible by 7, then 999 – 2×6 = 987 must be divisible by 7, and if 987 is divisible by 7, 98 – 2×7 = 84 must be divisible by 7, and we know that 84 is divisible by 7. Therefore, 999999 is divisible by 7.

This trick can be generalized to different numbers. For the mathematical minded: If you want to prove if a number 10X + Y is divisible by P, where X is a positive integer, and Y is an integer from 0 to 9 inclusive, and find a K such that 10K + 1 is divisible by P, then X – KY must also be divisible by P.

In layman’s terms: You know that the divisibility rule for 7 involves subtracting 2 times the last digit from the other digits. A similar trick can be applied to other odd numbers. The reason why the last digit is multiplied by 2 is because 21 is the least multiple of 7 that ends in a 1. The divisibility rule for 13 is similar, but you would have to subtract 9 times the second digit from the other digits, as 91 is the least multiple of 13 ending in a 1, and for 17, you would multiply the last digit by 5, as 51 is the least multiple of 17 ending in a 1.

Are you suspicious? I don’t blame you, but see for yourself. Multiply 17 by a large number on a calculator, and try the trick on the large number. For example, you can try 83521. 8352 – 5×1 = 8347, 834 – 5×7 = 799, 79 – 9×5 = 34, and 34 is divisible by 17, so 83521 must also be divisible by 17.

You can do this with any odd ending in 1, 3, 7, or 9, as all odds ending in these numbers will eventually have a multiple ending in 1. Try to find the divisibility rule for 97. Scroll down to see the answer when you find it.

Answer: Subtract 29 times the second digit from the first.