this post was submitted on 04 Nov 2024
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Science Memes

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[–] darkpanda@lemmy.ca 13 points 3 weeks ago (2 children)

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?

[–] TwilightKiddy@programming.dev 8 points 3 weeks ago (2 children)

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".

[–] darkpanda@lemmy.ca 9 points 3 weeks ago* (last edited 3 weeks ago) (1 children)

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 too large to fit in this comment.

[–] Scubus@sh.itjust.works 3 points 3 weeks ago (1 children)

Its never divisible by zero, and its always divisible by one

[–] wicked@programming.dev 3 points 3 weeks ago (1 children)
[–] LordTrychon@startrek.website 1 points 3 weeks ago

Interesting read. Thank you.

[–] levzzz@lemmy.world 1 points 3 weeks ago

The 9 rule works for 3 too The 6 rule is if (divisible by 3 and divisible by 2)