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And 299999999 is divisible by 13 (lemmy.world)

submitted 2 weeks ago by [email protected] to c/[email protected]

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[–] [email protected] 13 points 2 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 8 points 2 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".

[–] [email protected] 9 points 2 weeks ago* (last edited 2 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.

[–] [email protected] 2 points 2 weeks ago

Fucking lol

[–] [email protected] 3 points 2 weeks ago (1 children)

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

[–] wicked 3 points 2 weeks ago (1 children)

Never say never. 1/0 = 0

[–] [email protected] 1 points 2 weeks ago

Interesting read. Thank you.

[–] [email protected] 1 points 2 weeks ago

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