this post was submitted on 08 Jan 2025
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[–] testfactor@lemmy.world 36 points 1 day ago (2 children)

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.)

[–] brian@lemmy.ca 7 points 1 day ago (1 children)

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

[–] sudo@lemmy.today 7 points 1 day ago

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.

[–] GrammarPolice@lemmy.world 0 points 1 day ago (3 children)
[–] testfactor@lemmy.world 10 points 1 day ago (1 children)
[–] GrammarPolice@lemmy.world 3 points 1 day ago* (last edited 1 day ago)

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.

[–] Hjalamanger@feddit.nu 2 points 1 day ago

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 + 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.