Il do it for disability by three and a three digit numbers with the digits a, b and c. The value of that number then is 100a + 10b + c. They concept is the same for nine.
100a + 10b + a mod3 =
a + b + a
This means that, mod 3, a three digit number is equivalent to the sum of it’s digits and therefore preserves disability by 3.
Now provide the proof
https://brilliant.org/wiki/proof-of-divisibility-rules/
The 7 and 13 rules are pretty cool too.
This is insane stuff. 13 is truly mesmerizing. Although I don’t think I’m sharp enough for the proofs. Even the divisibility by 2 proof looks hellish.
Il do it for disability by three and a three digit numbers with the digits a, b and c. The value of that number then is 100a + 10b + c. They concept is the same for nine.
100a + 10b + a mod 3 = a + b + aThis means that, mod 3, a three digit number is equivalent to the sum of it’s digits and therefore preserves disability by 3.
I have discovered a truly marvelous demonstration of this proposition that this comment section is too narrow to contain.