aka Matěj Sál

  • I live in Andromeda Galaxy (nah. I live in Prague, Czech Republic)
  • I was born on October 9
  • I am useless
  • Unknown95387

    You probably already know it, but there are bigger roman numerals than just M = 1,000. You know that there are roman numerals that we know from school:

    Roman numeral Number
    I 1
    V 5
    X 10
    L 50
    C 100
    D 500
    M 1,000

    ... and that was all.

    There are bigger roman numerals. It's not written badly, like MMMMMMMMMM = 10,000 and so on. The bar on the top of the numeral means "times 1,000". In this case, 1,000 is also equal to \(\bar{\text{I}}\) . So it's :

    Roman numeral Number
    \(\bar{\text{V}}\) 5,000
    \(\bar{\text{X}}\) 10,000
    \(\bar{\text{L}}\) 50,000
    \(\bar{\text{C}}\) 100,000
    \(\bar{\text{D}}\) 500,000
    \(\bar{\text{M}}\) 1,000,000

    Combining numerals isn't hard.

    4,000 = \(\text{M}\bar{\text{V}}\) or \(\bar{\text{IV}}\)

    6,000 = \(\bar{\text{V}}\text{M}\) or \(\bar{\text{V…

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  • Unknown95387

    Help please

    November 26, 2017 by Unknown95387

    This is totally noob question for you, but I need to ask you something.

    I have an idea for a number called \(mag(3)\), where mag is a function. Mag is a function with \(n\) gag's, and \(gag(n)\) is equal to \(A(n,n)\) in Ackermann function (\(n\)'s instead of \(x\) and \(y\)). So there are 3 gag's to make up mag(3). The first \(gag(3)\) is 61. The second one is much much larger and I think it's equal to \(f_\omega(60)\) (Largest number I've ever got to, and we're not at the end !). Now there's the problem:

    I've tried many ways to compute the third \(gag\) and I ended with \(f_{\omega}(60)\{f_{\omega}(60)\{f_{\omega}(60)\}f_{\omega}(60)\}f_{\omega}(60)\) or \(f_{\omega}(60)\{A(f_{\omega}(60))\}f_{\omega}(60)\) with usage of Ackermann numbers.…

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  • Unknown95387

    I spent approximately 30 minutes doing this because of 1 single video (I think you've saw it somewhen before). The reason I calculated it is because I was bored.

    Of course there's a big chance that this cube wouldn't exist in real life, but nvm.

    So here's the calculation.

    \(\frac{8! \times 3^{7} \times 24!^{1,001,000}}{24^{5,940,151}} = 2.095127 \times 10^{23,816,506}\)

    I'm not sure if it's correct, but I kinda think it is.

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  • Unknown95387

    I have a question for everyone on this Wikia. How will we name numbers that comes after Millillion ? I found 2 sources:

    First is this webpage. There is millillion called milliatillion and trimilliaduotrigintatrecentillion is called tremilliatrecendotrigintillion. But then there's the problem. Both articles are here with same equal (109,999) (And that's the reason why am I asking to this question).

    The second source is Conway and Guy's naming system, where millillion is used. At the end of this video, we can se continue after millillion, which is millimillion, millibillion, millitrillion etc. I have made my page, where I'm writing numbers that are larger than millillion (And numbers with exponent that looks like 99,999,999...).

    So in which sys…

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  • Unknown95387

    I need help

    October 7, 2017 by Unknown95387


    Can I ask someone how did you added mathjax to this wiki ? Because I tried to use templates, but it doesn't work.

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