• sem@lemmy.blahaj.zone
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    2 hours ago

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

  • grrgyle@slrpnk.net
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    4 hours 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

  • Etterra@lemmy.world
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    4 hours 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?

  • m_f@midwest.social
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    11 hours 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
    

    Related to cyclic numbers:

    142857 * 1 = 142857
    142857 * 2 = 285714
    142857 * 3 = 428571
    142857 * 4 = 571428
    142857 * 5 = 714285
    142857 * 6 = 857142
    142857 * 7 = 999999
    
      • prime_number_314159@lemmy.world
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        31 minutes 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.

  • TwilightKiddy@programming.dev
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    16 hours 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.

  • OfficerBribe@lemm.ee
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    15 hours 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.

    And as for 13:

    Form the alternating sum of blocks of three from right to left. The result must be divisible by 13

    So we have 999 - 999 + 299 = 299.

    You can continue with other rules so we can then take this

    Add 4 times the last digit to the rest. The result must be divisible by 13.

    So for 299 it’s 29 + 9 * 4 = 65 which divides by 13. Pretty cool.

    • Grubberfly 🔮@mander.xyz
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      2 hours ago

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

      It should be fun to develop each proof. Particularly the 1,3,2,-1,-3,-2 rule, which at first sight seems could be easily expanded to any other number.

    • Shard@lemmy.world
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      16 hours ago

      Yes technically almost every number is divisible by another in some way and you’re left with a remainder that spans plenty of decimal places.

      But common parlance when something is said to be divisible is that the end results is a round number…