• logicbomb@lemmy.world
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    1 year ago

    Also, any number whose digits sum to a multiple of 3 is divisible by 3. For 51, 5+1=6, and 6 is a multiple of 3, so 51 can be cleanly divided by 3.

      • 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.

    • vortic@lemmy.world
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      1 year ago

      I’d forgotten this trick. It works for large numbers too.

      122,300,223÷3 = 40,766, 741

      1+2+2+3+2+2+3 = 15

      • ledtasso@lemmy.world
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        1 year ago

        This one has always bothered me a bit because …999999 is the same as infinity, so when you’re “proving” this, you’re doing math using infinity as a real number which we all know it’s not.

        • Snazz@lemmy.world
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          1 year ago

          You can also prove it a different way if you allow the use of the formula for finding the limit of the sum of a geometric series on a non-convergent series.

          Sum(ar^n, n=0, inf) = a/(1-r)

          So,

          …999999

          = 9 + 90 + 900 + 9000…

          = 9x10^0 + 9x10^1 + 9x10^2 + 9x10^3…

          = Sum(9x10^n, n=0, inf)

          = 9/(1-10)

          = -1

        • yetAnotherUser@feddit.de
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          1 year ago

          Yes, you’re right this doesn’t work for real numbers.

          It does however work for 10-adic numbers which are not real numbers. They’re part of a different number system where this is allowed.

  • Dagwood222@lemm.ee
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    1 year ago

    Any number where the individual digits add up to a number divisible by ‘3’ is divisible by 3.

    51 = 5+1 = 6, which is divisible by three.

    Try it, you’ll see it always works.