this post was submitted on 04 Nov 2024
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Science Memes

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[–] TwilightKiddy@programming.dev 76 points 15 hours ago (11 children)

The divisability rule for 7 is that the difference of doubled last digit of a number and the remaining part of that number is divisible by 7.

E.g. 299'999 → 29'999 - 18 = 29'981 → 2'998 - 2 = 2'996 → 299 - 12 = 287 → 28 - 14 = 14 → 14 mod 7 = 0.

It's a very nasty divisibility rule. The one for 13 works in the same way, but instead of multiplying by 2, you multiply by 4. There are actually a couple of well-known rules for that, but these are the easiest to remember IMO.

[–] darkpanda@lemmy.ca 11 points 15 hours ago (7 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 7 points 15 hours 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 6 points 15 hours ago* (last edited 12 hours 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 14 hours ago (1 children)

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

[–] wicked@programming.dev 2 points 7 hours ago (1 children)
[–] LordTrychon@startrek.website 1 points 2 hours ago

Interesting read. Thank you.

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