Comment on Panik

• Tangent5280@lemmy.world ⁨1⁩ ⁨year⁩ ago

  Our plan to find the witch has worked, boys! Get her!

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  • Karyoplasma@discuss.tchncs.de ⁨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.

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• match@pawb.social ⁨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

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• AccountMaker@slrpnk.net ⁨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?

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  • Colalextrast@lemmynsfw.com ⁨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.

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    • AccountMaker@slrpnk.net ⁨1⁩ ⁨year⁩ ago

      Oooh, I didn’t realize that you subtract from the original number without the last digit. Thanks

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  • Kwdg@discuss.tchncs.de ⁨1⁩ ⁨year⁩ ago

    Yeah you got it wrong, it’s 1 - 42 = -7 2 - 12 = 0 2 - 8*2 = -14

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• cactusupyourbutt@lemmy.world ⁨1⁩ ⁨year⁩ ago

  okay I understand that this works, but is there a mathematical proof for this?

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  • TauZero@mander.xyz ⁨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.

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  • Karyoplasma@discuss.tchncs.de ⁨1⁩ ⁨year⁩ ago

    math.stackexchange.com/…/proof-for-divisibility-b…

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