Yep. If the sun of the numbers is divisible by 3, the number is divisible by three.
Works great for 6 too, as if it’s divisible by 3 and even, the number is divisible by 6.
And 9 is the same thing, but the sum has to be divisible by 9 (e.g. 12384 is divisible by 9 because the sum of the digits is 18, which is divisible by 9)
There’s also good rules for 4 and 8 as well. If the last 2 digits are divisible by 4, the whole number is (e.g. 127924 is divisible by 4 because 24 is) and if the last 3 numbers are divisible by 8, the whole number is (e.g. 12709832 is divisible by 8 because 832 is.)
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 + amod3 =
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.
Yep. If the sun of the numbers is divisible by 3, the number is divisible by three.
Works great for 6 too, as if it’s divisible by 3 and even, the number is divisible by 6.
And 9 is the same thing, but the sum has to be divisible by 9 (e.g. 12384 is divisible by 9 because the sum of the digits is 18, which is divisible by 9)
There’s also good rules for 4 and 8 as well. If the last 2 digits are divisible by 4, the whole number is (e.g. 127924 is divisible by 4 because 24 is) and if the last 3 numbers are divisible by 8, the whole number is (e.g. 12709832 is divisible by 8 because 832 is.)
You just casually dropping in that 832 is divisible by 8 makes me feel as if there’s a small gap in our abilities to do mental math
832 is 800 + 32
800 is obviously divisible by 8, so it can also be negated like the first few digits. 32 is also divisible by 8.
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.
I have discovered a truly marvelous demonstration of this proposition that this comment section is too narrow to contain.
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.