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Panik (i.postimg.cc)
submitted 2 years ago by to c/lemmyshitpost@lemmy.world
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[–] 23 points 2 years ago (3 children)

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.

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  • [–] 16 points 2 years ago (1 child)

    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...

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  • [–] 18 points 2 years ago (3 children)

    Another way to tell if 59271 is divisible by 7 is to divide it by 7. It will take about the same amount of time as the trick you're presenting, and then you'll already have the result.

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  • [–] 5 points 2 years ago (1 child)

    I'm sure every digit has rules to figure it out if you get technical enough.

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