this post was submitted on 08 Jan 2025
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[–] GrammarPolice@lemmy.world 0 points 7 months ago (3 children)
[–] testfactor@lemmy.world 10 points 7 months ago (1 children)
[–] GrammarPolice@lemmy.world 3 points 7 months ago* (last edited 7 months 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.

[–] TempermentalAnomaly@lemmy.world 2 points 7 months ago

I have discovered a truly marvelous demonstration of this proposition that this comment section is too narrow to contain.

[–] Hjalamanger@feddit.nu 2 points 7 months 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.