• 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.

      • 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

      • 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?

        • 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.

  • 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.

  • 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.

  • TimewornTraveler@lemm.ee
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    1 year ago

    weird how ppl are getting all excited over this. weirder all the random math facts on the comments. and everyone checking with long math as if it might not be lol. I guess I’ll throw a few math facts in?

    17 is a prime number. 3 is a prime number.

    all numbers can be factored down to primes.

    19 is a prime number.

    19*3=57. is that one gross too?

  • dbilitated@aussie.zone
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    1 year ago

    I actually really like this. 17 is three less than 20, 20x3 is 60, 3x3 is 9, 51 is 60-9. It just feels nice how it all fits together.

  • MystikIncarnate@lemmy.ca
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    1 year ago

    Technically, isn’t everything divisible by any number? You just get remainders and/or fractions in the result?

    I mean, I still didn’t want to know this, but…

    • acutfjg@feddit.nl
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      1 year ago

      The definition for “divisible” is being able to be divided without a remainder.

  • 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.

    • Sadbutdru@sopuli.xyz
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      1 year ago

      Does this also work the other way round, i.e. do all multiples of three have digits that sum to a multiple of 3? All the ones I’ve checked so far do, but is it proven?

      • Goddard Guryon@sopuli.xyz
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        1 year ago

        Indeed, an integer is divisible by 3 if and only if the sum of its digits is divisible by 3.

        For proof, take the polynomial representation of an integer n = a_0 * 10^k + a_1 * 10^{k-1} + … + a_k * 1. Note that 10 mod 3 = 1, which means that 10^i mod 3 = (10 mod 3)^i = 1. This makes all powers of 10 = 1 and you’re left with n = a_0 + a_1 + … + a_k. Thus, n is divisible by 3 iff a_0 + a_1 + … + a_k is. Also note that iff answers your question then; all multiples of 3 have to, by definition, have digits whose sum is a multiple of 3

    • 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