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If all of the digits summed recursively reduce to a 9, then the number is divisible by 9 and also by 3.

If the difference between the sums of alternating sets digits in a number is divisible by 11, then the number itself is divisible by 11.

That’s all I can remember, but yay for math right?

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TwilightKiddy@programming.dev ⁨1⁩ ⁨week⁩ ago
Well, on the side of easy ones there is “if the last digit is divisible by 2, whole number is divisible by 2”. Also works for 5. And if you take last 2 digits, it works for 4. And the legendary “if it ends with 0, it’s divisible by 10”.

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- darkpanda@lemmy.ca ⁨1⁩ ⁨week⁩ ago
  There’s also the classic “no three positive integers a, b, and c to satisfy a**n + b**n = c**n for values of n greater than 2“ trick but my proof is to large to fit in this comment.
  
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  - JamesStallion@sh.itjust.works ⁨1⁩ ⁨week⁩ ago
    Fucking lol
    
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- Scubus@sh.itjust.works ⁨1⁩ ⁨week⁩ ago
  Its never divisible by zero, and its always divisible by one
  
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  - wicked@programming.dev ⁨1⁩ ⁨week⁩ ago
    Never say never. 1/0 = 0
    
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    LordTrychon@startrek.website ⁨1⁩ ⁨week⁩ ago
    Interesting read. Thank you.
    
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levzzz@lemmy.world ⁨1⁩ ⁨week⁩ ago
The 9 rule works for 3 too The 6 rule is if (divisible by 3 and divisible by 2)

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