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And 299999999 is divisible by 13 (lemmy.world)

submitted 1 month ago by [email protected] to c/[email protected]

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[–] [email protected] 13 points 1 month 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?

[–] [email protected] 8 points 1 month 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".

[–] [email protected] 9 points 1 month ago* (last edited 1 month 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.

[–] JamesStallion 2 points 1 month ago

Fucking lol

[–] Scubus 3 points 1 month ago (1 children)

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

[–] [email protected] 3 points 1 month ago (1 children)

Never say never. 1/0 = 0

[–] [email protected] 1 points 1 month ago

Interesting read. Thank you.

[–] [email protected] 1 points 1 month ago

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