And 299999999 is divisible by 13

Comments

• TwilightKiddy@programming.dev ⁨1⁩ ⁨week⁩ 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.

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  • urda@lebowski.social ⁨1⁩ ⁨week⁩ ago

    Thanks I hate it

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    • veroxii@aussie.zone ⁨1⁩ ⁨week⁩ ago

      This math will not stand man!

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  • darkpanda@lemmy.ca ⁨1⁩ ⁨week⁩ 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?

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    • TwilightKiddy@programming.dev ⁨1⁩ ⁨week⁩ 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”.

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    • levzzz@lemmy.world ⁨1⁩ ⁨week⁩ ago

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

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  • candybrie@lemmy.world ⁨1⁩ ⁨week⁩ ago

    I think it might be easier just to do the division.

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• m_f@midwest.social ⁨1⁩ ⁨week⁩ ago

  ⅐ = 0.1̅4̅2̅8̅5̅7̅

  The above is 42857 * 7, but you also get interesting numbers for other subsets:

       7 * 7 =     49
      57 * 7 =    399
     857 * 7 =   5999
    2857 * 7 =  19999
   42857 * 7 = 299999
  142857 * 7 = 999999
  

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• Usernamealreadyinuse@lemmy.world ⁨1⁩ ⁨week⁩ ago

  42857 for those who wonder

  And for ops title: 23076923

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• AI_toothbrush@lemmy.zip ⁨1⁩ ⁨week⁩ ago

  Actually disgusting

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  • ryannathans@aussie.zone ⁨1⁩ ⁨week⁩ ago

    Why

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• OfficerBribe@lemm.ee ⁨1⁩ ⁨week⁩ ago

  Never realized there are so many rules for divisibility. This post fits in this category:

  Forming an alternating sum of blocks of three from right to left gives a multiple of 7

  299,999 would be 999 - 299 = 700 which is divisible by 7. And if we simply swap grouped digits to 999,299, it is also divisible by 7 since 299 - 999 = -700.

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  • grubberfly@mander.xyz ⁨1⁩ ⁨week⁩ ago

    That is indeed an absurd amount of rules (specially for 7) !

    It should be fun to develop each proof. Specially the 1,3,2,-1,-3,-2 rule

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• Revan343@lemmy.ca ⁨1⁩ ⁨week⁩ ago

  Thanks, Satan

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  • Klear@lemmy.world ⁨1⁩ ⁨week⁩ ago

    Wait until you leard of 51/17.

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    • Revan343@lemmy.ca ⁨1⁩ ⁨week⁩ ago

      Thanks, I hate it

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    • prime_number_314159@lemmy.world ⁨1⁩ ⁨week⁩ ago

      With 17, I understand that you’re referring to how 299,999 is also divisible by 17. What is the 51 reference, though? I know there’s 3,999,999,999,999 but that starts with a 3. Not the same at all.

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• chemicalwonka@discuss.tchncs.de ⁨1⁩ ⁨week⁩ ago

  I know you opened your calculator app to check it.

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• SerpentPeaked@lemmynsfw.com ⁨1⁩ ⁨week⁩ ago

  Isn’t every number divisible by 7?

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  • ilovededyoupiggy@sh.itjust.works ⁨1⁩ ⁨week⁩ ago

    Yup, this is just a coordinated smear campaign from Big Integer.

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• Etterra@lemmy.world ⁨1⁩ ⁨week⁩ ago

  So what? Being a prime number doesn’t mean it can’t be a divisor. Or is it the string of 9s that’s supposed to be upsetting? Why? What difference does it make?

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• Grandwolf319@sh.itjust.works ⁨1⁩ ⁨week⁩ ago

  49 is divisible by 7, so why not?

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• grrgyle@slrpnk.net ⁨1⁩ ⁨week⁩ ago

  Can we just say it isn’t? Like that’s an exception, and then the rest of math can just go on like normal

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• sem@lemmy.blahaj.zone ⁨1⁩ ⁨week⁩ ago

  Phew, for a moment I worried that 2.9999… was divisible by 7 and I woke up in some kind of alternate universe

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• henfredemars@infosec.pub ⁨1⁩ ⁨week⁩ ago

  …9999 is exactly equal to -1.

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  • bstix@feddit.dk ⁨1⁩ ⁨week⁩ ago

    But only in base 10.

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    • henfredemars@infosec.pub ⁨1⁩ ⁨week⁩ ago

      Well yes of course. If it was a different base, writing it that way if the symbol was even available would be a different number.

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