• logicbomb@lemmy.world
    link
    fedilink
    arrow-up
    23
    ·
    1 year ago

    I only know rules for 2 (even number), 3 (digits sum to 3), 4 (last two digits are divisible by 4), 5 (ends in 5 or 0), 6 (if it satisfies the rules for both 3 and 2), 9 (digits sum to 9), and 10 (ends in 0).

    I don’t know of one for 7, 8 or 13. 11 has a limited goofy one that involves seeing if the outer digits sum to the inner digits. 12 is divisible by both 3 and 4, so like 6, it has to satisfy both of those rules.

    • beckerist@lemmy.world
      link
      fedilink
      arrow-up
      16
      ·
      1 year ago

      7 is double the last number and subtract from the rest

      749 (easily divisible by 7 but for example sake)

      9*2=18

      74-18=56

      6*2=12

      5-12= -7, or if you recognize 56 is 7*8…


      I’ll do another, random 6 digit number appear!

      59271

      1*2=2

      5927-2=5925

      5*2=10

      592-10=582

      2*2=4

      58-4=54, or not divisible

      I guess for this to work you should at least know the first 10 times tables…

    • octoperson@sh.itjust.works
      link
      fedilink
      arrow-up
      4
      ·
      1 year ago

      11 is alternating sum
      So, first digit minus second plus third minus fourth…
      And then check if that is divisible by 11.