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THE MATH OF THE CHARLOTTE FAMILY

Discussion in 'The Theory Archive' started by Icecream001, Jun 12, 2016.

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  1. Icecream001

    Icecream001

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    Oi! This was done out of boredom. I've never seen anyone do this, and now I'm tired of waiting. So I think I figured out the Charlotte family. I suck at math, but I'm suprised that this shit works. Ok prepare to think. You might not understand a thing I'm saying but, just TRY:

    (keep in mind even though I barely use this)
    S=SONS
    H=HUSBANDS
    d=daughters

    #=Number

    Here's my prediction: ALL HUSBANDS, EXCEPT FOR 5, HAD A SON AND A DAUGHTER.
    What we know:
    There are 39 daughters, 46 sons, 43 husbands and one mother. The total is: 129 people

    My method for the SON part is trying to get the # of sons = to the # of husbands and see how many EXTRA sons there are.


    Here's what I'm doing:
    S: 46-3= 43 sons. Now that I made the # of sons equal to the # of husbands, I came to this conclusion:

    If there were never 3 extra sons and instead there were 43 sons, that means that every husband had 1 son, if they never had any daughters. BUT there are 2 possibilities:

    1 H could've had 4 S
    or
    3 H had 2 Sons each

    We'll see which one works later. I'll put everything together at the end.

    My method for the DAUGHTER part is a bit more different and complex. Here's what I am doing:

    39 Daughters right?
    43 HUSBANDS - 4 = 39

    If there were never 4 extra husbands then each H would have 1 DAUGHTER, if they never had sons. BUT there's a problem:

    LOLA and that lady in Ch 825 are probably twins. So 1 husband had more than 1 daughter so there has to be 38 husbands in order for that to work. SO THERE ARE ACTUALLY 5 EXTRA HUSBANDS.

    So 38 husbands had 39 daughters. Wait, but how is it possible for the # of Daughters to be more than # of husbands? Well like I said, there is ONE set of twins. ONLY ONE. Why?! because if they never had twins then subtract 2 from the # of daughters.

    That also means that there would be 1 husband out. Here's what I mean:
    39 daughters - 2 daughters(twins) = 37 daughters
    Only 1 husband could have had 2 twins so: 38 husbands - 1 husband = 37 husbands

    37 = 37. The #'s match. But what about the rest of the Husbands. 43 - 38 = 5 husbands. So there are 5 husbands left which I mentioned before.

    OK LET'S PUT IT TOGETHER AND SEE:
    I'm going to use the lowest # of siblings. The Daughters. I will try and make the sons equal to 39. For husbands, they will equal to 38. and whatever I subtracted to do that, will be the # of husbands left. Same for the sons.

    43-5=38 husbands. 5 Husbands left
    46-7=39 sons. 7 sons left.

    5 husbands left.
    Remember the possiblities in the Son part? Well, the only onne that works is that 3 husbands had 2 sons each. The reason is because we need to get to 43 husbands and 46 sons in order for this to work. So if there are 3 more husbands, then:
    38+3=41

    Now there are 2 more husbands left. One husband is the husband that gave birth to the twins of the daughters. The LAST husband I'm getting to.

    In total:
    41+2=43 husbands.
    WE DID THE HUSBAND PART! 3 husbands each had 2 sons, 1 husband had twins(2 daughter), and 1 husband is left. Now for the SONS:


    If 3 H had 2 sons each, then that means 6 sons were born. Remember when we did this: 46 sons - 7=39 sons. That means there is 1 son missing. THIS SON WILL BE PUT WITH THE LAST HUSBAND I MENTIONED. SO: 7 extra sons + 39 sons = 46 sons in all

    Now, this part I didn't plan out but, using the numbers I have, every son had to have a sister EXCEPT for 7 sons. That means that, every husband had a son and a daughter, EXCEPT for 5. Those 5 husbands. 3 H each had 2 sons each, 1 had twin daughters, and the last one had only 1 son.

    So now that that's done, my experiment is complete. Wait! but who's the middle child?? Well, the total siblings is 85. Since this # is odd then there has to be a # in the middle. So what I did was find the middle child by doing this:

    46 - 39 = 7. There are seven numbers between 46 and 39. They are
    1=45 2=44 3=43 4=42 5=41 6=40 7=39

    THE 42ND SIBLING IS THE MIDDLE CHILD. Its hard to decide if it is a daughter or a son. But I did this:
    SON - DAUGHTER = SAME DIFFERENCE
    45-38=7
    44-37=7
    43-36=7

    42-35=7
    40-33=7
    41-34=7
    39-32=7


    Note: There's actually a possibility that this 42nd son can be the son that doesn't have any brothers or sisters and his dad is the husband that has only 1 son. That would mean that the 35th daughter doesn't have a dad.

    It could be that the 42nd son or the 35th daughter is the middle child but it depends on when they were born. I think the son would be born first..............:neutral:

    Speculation:
    HOLD ON. WAIT!@ OH SH!T! ISN'T PUDDING THE 35TH DAUGHTER. WELL FUCK, PUDDING MIGHT BE THE MIDDLE CHILD THEN. AND THE MIDDLE CHILD IS ALWAYS THE FUCKING ONE IN THE MIDDLE OF EVERYTHING. I KNOW HOW IT FEELS, I'M THE MIDDLE CHILD TOO. I KNEW THERE WAS SOMETHING SPECIAL ABOUT PUDDING. ALSO THERE'S A POSSIBILITY THAT PUDDING HAS A BROTHER. But, since ONE of the husbands, HAVE to have only 1 child, which is a son...........PUDDING MIGHT BE ADOPTED!!!!!!!


    Anyways, I hope your brain is okay. I did this out of boredom and I suck at math, but it fucking worked. I hope you understand everything. I got math finals this week too but Thank you for reading if you did.
     
    Last edited: Jun 12, 2016
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  2. Quezon D. Faizul

    Quezon D. Faizul

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    what is mean by '#'..?

    *its actually damn complicated.lol

    * pudding is adopted daughter..? it becames more complicated tbh,xD
     
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  3. Icecream001

    Icecream001

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    # means number
     
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  4. Mirai

    Mirai

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    Good work but a bit complicated for em :D
     
  5. Shanksette

    Shanksette

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    I didn't read everything but, from what I understand, there are 85 children born from 43 husbands, if we suppose she doesn't do 'it' twice with the same man, it corresponds to around 33 years of activity (assuming she follows human biology and has 9 months-pregnancy). So there's an average of 2 children per pregnancy cycle, which means to every single child there is a complementary triplet (or there is a quadruplet for two single children, or Big Mom re-used some husbands, which would increase the number of pregnancy cycles and decrease the number of twins). Let's say Big Mom can't bear more than 10 children at a time in her belly (I think the world record is 6 or something like that). It's a pretty safe bet to assume that each husband had at least one child. So I think it can go from your theory (as equally divided as possible) to the situation where 4 husbands have 10 children each, one has 7 and the other ones all have one child.
     
  6. Watermelon

    Watermelon

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    39 daughters, 46 sons, 43 husbands

    38 husbands each had 1 son and 1 daughter.
    1 husband had twin daughters.
    3 husbands each had 2 sons.
    1 husband had 1 son.

    If this is correct then there would be 40 daughters.
     
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  7. Hi Shin Unit

    Hi Shin Unit

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    I think its 8 or 9 for the world record.
    Edit what I search on google.. It says 9*
    http://www.guinnessworldrecords.com...ldren-delivered-at-a-single-birth-to-survive/
    Boy oh boy.. I was wrong... NEW Record 17 babies born after 29 hours.
    http://worldnewsdailyrepor****m/usa-mother-gives-birth-to-17-babies-at-once/
     
    Last edited: Jun 12, 2016
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  8. bronion

    bronion

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    Nobody told me you need math in One Piece! :DD

    I'm out.
     
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  9. Shanksette

    Shanksette

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    I would trust the guinness official figure rather than the other one :D the comments say that it's fake x)

    In any case, there are a lot of incoherences in the original post, I'll post a point-by-point analysis later.
     
  10. Xlaw

    Xlaw

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    Well...I don't understand why people r so surprised by BM's family tree...
    In real life, man name Ziona chan has 39 wives , 94 wives and 33 grandchildrens...

    Also Feodore Vesielv (Russia) and his "one wife" gave birth to 69 children's (16 twins, 7 time - 3 at once, 4 time- 4 at once) and second wife had 18 children....means 82 in total....

    And this is manga so it can possible lol..
     
  11. Shanksette

    Shanksette

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    Ok, here we go. @Icecream001 I’m gonna be harsh, but please keep in mind that I just want to help :)

    Before anything else, this:
    43-5=38. According to you, 38 H had 1 S and 1 D each, 3 H had 2 S each, 1 H had 2 D, and 1 H had 1 S. That totals to:
    • 38+3+1+1=43 H -> ok
    • 38+2*3+1=45 S -> NO!
    • 38+2=40 D -> NO!

    All right, don’t panic, let’s follow your reasoning step by step and see where you made mistakes.

    Usually this is a sign that you should double check your solution :D

    I agree. Let’s assume one thing: each husband has at least one child. Why can we make this assumption? Because there are more children than husbands. Why do we make this assumption? Because Big Mom would probably have gotten rid of any husband that couldn’t give her a child (this is hypothetical).
    Now, if the husbands never had any daughters, and if there were 43 sons to 43 husbands, then the only solution would be for each husband to have 1 son. But there are daughters and there are more than 43 sons, so the solution we just found doesn’t work, since it’s the solution to another problem than the one we’re facing.

    There’s also another possibility: 1H has 2S, 1H has 3S, and the others have one son each. This is IN THE CASE where there are no daughters, which is NOT the case of the Big Mom family problem. In the Big Mom family problem, there are lots of other possibilities.

    Good remark here. If we assume that Lola and Lolookalike are indeed twins, then the husband who gave birth to them had at least 2 daughters, which leaves 37 other daughters. Therefore, there can’t be more than 38 husbands who had daughters. Note that this would be true even if Lola and Lolookalike were not strict twins, they can simply be sisters (i.e. have the same father). However, we do not know if there are exactly 5 extra husbands, there could be more. We do not know if Lola’s father had other daughters beside her sister and herself, and we do not know if there are other pairs of twin daughters.

    I give up trying to understand this sentence :lmao:

    No. The possibilities you found in the sons part were under the assumption that there were no daughters. But it’s NOT THE SAME problem as the Big Mom family problem, so your solution DOES NOT apply here. For example, a perfectly possible situation would be:
    Husband #1234567891011121314151617181920212223242526272829303132
    Sons108432222111111111111100000000000
    Daughters01800010000000000011111111111111
    Husband #3334353637383940414243
    Sons00000000000
    Daughters11111111124
    Actually, you already counted the husband who had twin daughters, he’s one of the 38 (38 husbands for 39 daughters, remember? That is to say 37 who had one daughter each and one who had two daughters).

    From what I get, you’re trying to reach the following configuration:
    Husband #12345678943Total
    Sons112221111146
    Daughters000002111139
    Total112223222285
    I think what you tried to do was distributing the children as evenly as possible. It doesn’t necessarily have to be that way, it’s even not very probable that the distribution is so smooth. In real life there would be bumps, like triplets for example, or one husband having 6 children…

    Okay so basically what you did was:
    • you ordered the kids with the sons first and then the daughters, so there are three groups: 39 sons, then 7 sons, then 39 daughters,
    • the sons in the 2nd group have numbers 39-40-41-42-43-44-45, the middle one is the 4th in the 2nd group aka the 42nd person,
    • but since they weren’t ordered by age, all you could deduce was that the 42nd person is the middle child but you can’t say if it’s a guy or a girl
    Right?
    Two things: the middle child is not the 42nd one, it’s the 43rd one, and there was a simpler way to see it. Your mistake was to count the 39th child among the 7 people in the middle, whereas you should have started from 40, and gone up to 46 instead of stopping at 45.
    The simpler way is the following: there are 85 children, the middle child has as many younger siblings as older siblings. Him/her excepted, there are 84 children. So 42 of those are younger than the middle child, and the 42 others are older. Therefore the middle child is the 43th child (and not the 42nd).

    I don’t see where you’re getting at.

    What? :lmao:

    This is true. But statistically, the 42nd son and the 35th daughter are likely to be among the youngest ones (they were among the latest to arrive, there would have to be 41 sons born before the middle child AND almost all the daughters born after him if we wanted the 42nd son to be the middle child… It’s possible but not likely).

    Lol what the fuck bro :eek:

    Yeah pretty good thx

    Not really :lmao:

    No :dead:

    Good luck!! Try not to make too complicated reasonings :)

    No probs, I hope I didn’t offend you ^^
     
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  12. Icecream001

    Icecream001

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    CAN A MOD LOCK THE THREAD PLEASE? AT LEAST I TRIED TO ATTEMPT THIS COMPLICATED MATH, OBVIOUSLY FAILING, BUT AT LEAST I HAD THE BALLS TO MAKE A THREAD AND I EMBARRASSD MYSELF BECAUSE @Shanksette OVER HERE IS OBVIOUSLY A MATHEMATICIAN @blaze @Chris Mic
     
    Last edited: Jun 14, 2016
  13. Soma

    Soma

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    Interesting mathematical formula, Charlotte Family though, Big Mom must've done it... whoa...
     
  14. Aru

    Aru

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    Damn you, numbers!!! This must be @karofabulous curse spreading! Math everywhere! u__U

    Interesting thread, though! Nice work, mate!
     
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