• 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.

  • Eager Eagle@lemmy.world
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    1 year ago

    51 = 3*17

    3*17 = 17 + 17 + 17

    17 + 17 + 17 = (10+7) + (10+7) + (10+7)

    (10+7) + (10+7) + (10+7) = 30 + 21

    30 + 21 = 51

    yup, math checks out

    • _dev_null@lemmy.zxcvn.xyz
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      1 year ago

      I think you skipped a step:

      1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1

      Edit: Ohhhh, math by tens, I totally missed it. In that case, my mind wants to break it down to (10 * 5) + 1, and I’d still totally miss 17 as a possible factor.

      • driving_crooner@lemmy.eco.br
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        1 year ago

        You miss a couple os steps too.

        First, lets define the axioms, we’re using Peano’s for this exercise.

        Axiom 1: 0 is a natural number.

        Jump to axiom 6, define the succession function s(n) where s(n) = 0 is false, and for brevity s(0) = 1, s(s(0)) = 2 and so on…

    • BOMBS@lemmy.world
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      1 year ago

      51 = 3*17

      3*17 = 3 + 3 + 3 + 3 + 3 + 3 + 3 + 3 + 3 + 3 + 3 + 3 + 3 + 3 + 3 + 3

      3 + 3 + 3 + 3 + 3 + 3 + 3 + 3 + 3 + 3 + 3 + 3 + 3 + 3 + 3 + 3 = (2+1) + (2+1) + (2+1) + (2+1) + (2+1) + (2+1) + (2+1) + (2+1) + (2+1) + (2+1) + (2+1) + (2+1) + (2+1) + (2+1) + (2+1) + (2+1) + (2+1)

      (2+1) + (2+1) + (2+1) + (2+1) + (2+1) + (2+1) + (2+1) + (2+1) + (2+1) + (2+1) + (2+1) + (2+1) + (2+1) + (2+1) + (2+1) + (2+1) + (2+1) = 34 + 17

      34 + 17 = 51

      👌

  • theneverfox@pawb.social
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    1 year ago

    This is why I love the number 7. It’s the first real prime number. All the others are “first”…1?2?3?5? No, those aren’t prime numbers, they’re “first” in a long line of not-prime numbers.

    Then you get to 7. Is 27943 divisible by 7? If you take away 3 is it? If you add 4 is?

    I have no clue, give me 10 minutes or a calculator is the only answer

    That’s what a real prime number is.

    • Karyoplasma@discuss.tchncs.de
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      1 year ago

      Take the last digit of the number, double it and subtract it from the rest. If that new number is divisible by 7, the original one is as well. For your example:

      2794 - 6 = 2788

      I know 2800 is divisible by seven, so 2788 is not. Thus 27943 is not divisible by 7.

      Quick maff shows that neither subtracting 3 or adding 4 will make the original number divisible by 7. Adding 1 or subtracting 6 will tho.

        • Karyoplasma@discuss.tchncs.de
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          1 year ago

          For divisibility by 13, take the last number, multiply by 4 and add to the rest.

          For divisibility by 17, take the last number, multiply by 5 and subtract from the rest.

          For divisibility by 19, take the last number, multiply by 2 and add to the rest.

          In fact, you can adapt the method to check for divisibility by any prime number k.

      • Match!!@pawb.social
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        1 year ago

        Quick check for divisibility: subtract 7 from it. If the new number is divisible by 7, then the original number is too

        • TauZero@mander.xyz
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          1 year ago

          There is a mathematical algorithm that proves this works in all cases. However this rule is not actually all that impressive as it appears at first glance! The number of operations (comparisons/subtractions/multiplications) you need to do is equivalent to just long-dividing the number by 7.

          Consider: each operation of the rule removes one digit from the end. But you could just as easily apply the rule like “If the first digit is >=7, subtract 7 from it. Else, subtract the biggest multiple of 7 that will fit from the first two digits.” To skip multiplying, you can use the following jump table: if the first digit is 6, subtract 54 from the first 2 digits, if 5 subtract 49, if 4: 35, if 3: 28, if 2: 14, if 1: 07. That will also remove one digit from the front! But now you are just doing long division.

      • AccountMaker@slrpnk.net
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        1 year ago

        But what about 14, 21 and 28?

        14 - 4*2 = 6, not divisible by 7

        21 - 1*2 = 19, not divisible by 7

        28 - 8*2 = 12, not divisible by 7

        Or did I misunderstand the algorithm?

        EDIT: I didn’t realize that you remove the last digit when subtracting, got corrected in the replies.

        • Colalextrast@lemmynsfw.com
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          1 year ago

          It goes like this

          1. create 2 distinct numbers by isolating the last digit from the other. For example, 154 becomes 15 and 4.

          2. double the number derived from the last digit. So, the four becomes 8.

          3. subtract from the number derived from the preceeding digits. 15 - 8.

          4. the resulting number is 7. Seven is divisible by 7, so we know 154 is divisible by 7.

    • saigot@lemmy.ca
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      1 year ago

      27943 - 7*1000 = 20943

      20943 -7*3*1000 = 20943 - 21000 = -57

      -57 is not divisible by 7 therefore 27943 is not divisible by 7.

      • theneverfox@pawb.social
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        1 year ago

        The other posters algorithm was better, but I was exaggerating - ultimately my point is you have to math it out

  • 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.

    • letsgo@lemm.ee
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      1 year 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.

      • Dagwood222@lemm.ee
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        1 year 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.

        • steeznson@lemmy.world
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          1 year 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.

        • Zacryon@feddit.de
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          1 year ago

          I guess I was one of the lucky few who learned this in elementary school. And later again.

      • AEsheron@lemmy.world
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        1 year 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.

      • 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.

        • 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.

        • 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

  • forrgott@lemm.ee
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    1 year ago

    I love how every reply has like the opposite energy to the meme. I also find math to be generally awesome.

  • KSP Atlas@sopuli.xyz
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    1 year ago

    When you start playing modded minecraft you get really good at multiplying and dividing by 144

  • kingthrillgore@lemmy.ml
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    1 year ago

    I used to do this thing where I would figure out if a number was prime or not and it kept me sane. Realizing this isn’t, may have just caused my whole world to fall apart.