• logicbomb@lemmy.world
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      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
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        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
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        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.