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Comment on And 299999999 is divisible by 13
TwilightKiddy@programming.dev 1 year ago
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 1 year ago
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?
levzzz@lemmy.world 1 year ago
The 9 rule works for 3 too The 6 rule is if (divisible by 3 and divisible by 2)
TwilightKiddy@programming.dev 1 year ago
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 1 year ago
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 to large to fit in this comment.
JamesStallion@sh.itjust.works 1 year ago
Fucking lol
candybrie@lemmy.world 1 year ago
I think it might be easier just to do the division.