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 + 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.
Divisibility by 3 rule is real. If the sum of the digits of a number is divisible by 3, then the number itself is also divisible by 3. Same goes with 9. There’s an 11 rule, but it’s a bit convoluted.
51 --> 5+1 = 6, 6 is divisible by 3. This means that 51 is divisible by 3.
60 is divisible by 3, 60/3 = 20.
51 is 9 less than 60. 9 is divisible by 3. 9/3 = 3.
20 - 3 = 17.
Is this a real divisibility rule?
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.
Divisibility by 3 rule is real. If the sum of the digits of a number is divisible by 3, then the number itself is also divisible by 3. Same goes with 9. There’s an 11 rule, but it’s a bit convoluted.
There’s also a rule 34, but it’s super advanced.
Yes :)