• @[email protected]
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    3011 months 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.

    • @[email protected]
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      911 months ago

      There are tricks like that for a lot of numbers. For 7, chop off the last digit, double it and add it to what’s left. Repeat as required. If the result is divisible by 7 then the original number was. eg: 356 -> 35+12=47 not db7. 357 =>35+14 both db7 so we don’t even need to do the add.

      • @[email protected]
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        711 months ago

        They didn’t teach stuff like this in school, which is silly. This is the kind of thing that a kid would eat up. It’s like they wanted to make sure people hated math.

        • @[email protected]
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          411 months ago

          My experience of maths in high school was being taught a trick or method to solve a really specific type of problem every week. Sometimes the method would build off something we’d learnt the previous week.

          The whole thing was bottom-up learning where you get given piecemeal nuggets of information but never see the big picture. They completely lost me at around the age of 15. I eventually came back to maths later in life after studying formal logic in my philosophy undergrad degree.

      • @[email protected]
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        11 months ago

        Technically it does work for 6, more literally, still aiming for 3, not 6. That’s half of it, if the starting number is even and divisible by 3 then it is also divisible by 6.