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And 299999999 is divisible by 13 (lemmy.world)

submitted 3 days ago by [email protected] to c/[email protected]

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[–] [email protected] 13 points 3 days 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 8 points 3 days 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 3 days ago* (last edited 3 days 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.

[–] [email protected] 2 points 3 days ago

Fucking lol

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

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

[–] wicked 3 points 3 days ago (1 children)

Never say never. 1/0 = 0

[–] [email protected] 1 points 3 days ago

Interesting read. Thank you.

[–] [email protected] 1 points 3 days ago

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