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Stupid Math Tricks: Shuffling Water Molecules

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Image by Johnny Blood, via WikiCommons.

So, I was having some fun today reading through the amazing-but-true thread at Reddit, when I had the idea to mash some crazy things together. Follow me carefully on this (and feel free to correct my math if you notice mistakes):

There are about 7 billion (7×10^9) people on Earth.

If you replaced every person with another Earth full of people (49 billion-billion, or 4.9×10^19), you’d still have fewer people than there are molecules in a drop of water (divide the number of people by Avogadro’s number to get about 8.1368×10^-5 moles; there are 18mL of water in a mole of water, so a we get a little over .0014 mL of water; a drop of water has been defined in some places as 0.05mL).

Extrapolating back out, we can figure out that there are 3.346×10^28 molecules of water in a cubic meter of water (1×10^6 gms of water per m3, 18 grams per mole, Avogadro’s number again; everything is assuming standard temperature and pressure). The recognized volume of our sun is 1.4×10^27 cubic meters. Multiply those together to find out that there would be 4.684×10^55 molecules in a volume of water equal to that of our sun. That’s a big freakin’ number.

But…

It would take enough water to fill 1.7 TRILLION suns to have as many molecules of water as there are ways to shuffle a standard deck of cards (52! = 8.067×10^67). That’s more than 4 times the number of stars in the Milky Way galaxy.

::head asplode::

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6 thoughts on “Stupid Math Tricks: Shuffling Water Molecules

  1. Everybody.

    (N).
    z^3=[z(z+1)/2]^2 – [z(z-1)/2]^2

    With two any integers (a) and ( b)::
    [a(a+1)/2]^2 – [b(b+1)/2]^2=(b+1)^3+(b+2)^3+……..+a^3.
    Therefore:::
    z(z+1)/2]^2 =[6(6+1)/2]^2+(6+1)^3+(6+2)^3+……..+z^3
    [x(x+1)/2]^2=[5(5+1)/2]^2+(5+1)^3+(5+2)^3+……..+x^3.
    [y(y+1)/2]^2=[2(2+1)/2]^2+(2+1)^3+(2+2)^3+……..+y^3.
    And:
    :[z(z-1)/2]^2=[6(6+1)/2]^2+(6+1)^3+(6+2)^3+……..+(z-1)^3.
    [x(x-1)/2]^2=[5(5+1)/2]^2+(5+1)^3+(5+2)^3+……..+(x-1)^3
    [y(y-1)/2]^2=[2(2+1)/2]^2+(2+1)^3+(2+2)^3+……..+(y-1)^3

    Therefore, first equation:
    [z(z+1)/2]^2 – [x(x+1)/2]^2 – [y(y+1)/2]^2=[6(6+1)/2]^2+(6+1)^3+(6+2)^3+……..+z^3 – [5(5+1)/2]^2-(5+1)^3-(5+2)^3-……..-x^3 – [2(2+1)/2]^2-(2+1)^3-(2+2)^3-……..-y^3.
    And second equation:
    [z(z-1)/2]^2 – [x(x-1)/2]^2 – [y(y-1)/2]^2=[6(6+1)/2]^2+(6+1)^3+(6+2)^3+……..+(z-1)^3 – [5(5+1)/2]^2-(5+1)^3-(5+2)^3-……..-(x-1)^3 – [2(2+1)/2]^2-(2+1)^3-(2+2)^3-……..-(y-1)^3

    Because have an extra equation :
    [6(6+1)/2]^2 – [5(5+1)/2]^2 – [2(2+1)/2]^2.=[5(5+1)/2]^2 – 2[2(2+1)/2]^2
    Therefore, first equation become:
    [z(z+1)/2]^2 – [x(x+1)/2]^2 – [y(y+1)/2]^2=[5(5+1)/2]^2 – 2[2(2+1)/2]^2+(6+1)^3+(6+2)^3+……..+z^3 -(5+1)^3-(5+2)^3-……..-x^3 – (2+1)^3-(2+2)^3-……..-y^3.
    And second equation become:
    [z(z-1)/2]^2 – [x(x-1)/2]^2 – [y(y-1)/2]^2=[5(5+1)/2]^2 – 2[2(2+1)/2]^2+(6+1)^3+(6+2)^3+……..+(z-1)^3 -(5+1)^3-(5+2)^3-……..-(x-1)^3 – (2+1)^3-(2+2)^3-……..-(y-1)^3

    Having:
    [5(5+1)/2]^2 – 2[2(2+1)/2]^2=(-1)^3+(-2)^3+(2+1)^3+(2+2)^3+……..+5^3.
    Therefore:
    First equation:
    [z(z+1)/2]^2 – [x(x+1)/2]^2 – [y(y+1)/2]^2=(-1)^3+(-2)^3+(2+1)^3+(2+2)^3+……..+5^3+(6+1)^3+(6+2)^3+……..+z^3 -(5+1)^3-(5+2)^3-……..-x^3 – (2+1)^3-(2+2)^3-……..-y^3.
    And second equation
    [z(z-1)/2]^2 – [x(x-1)/2]^2 – [y(y-1)/2]^2=(-1)^3+(-2)^3+(2+1)^3+(2+2)^3+……..+5^3+(6+1)^3+(6+2)^3+……..+(z-1)^3 -(5+1)^3-(5+2)^3-……..-(x-1)^3 – (2+1)^3-(2+2)^3-……..-(y-1)^3

    Define the function:
    f(z,x,y)=[z(z+1)/2]^2 – [x(x+1)/2]^2 – [y(y+1)/2]^2
    So:
    f(z-1,x-1,y-1)=[z(z-1)/2]^2 – [x(x-1)/2]^2 – [y(y-1)/2]^2.
    And define another function:
    g(z,x,y)=(-1)^3+(-2)^3+(2+1)^3+(2+2)^3+……..+5^3+(6+1)^3+(6+2)^3+……..+z^3 -(5+1)^3-(5+2)^3-……..-x^3 – (2+1)^3-(2+2)^3-……..-y^3.
    So:
    g(z-1,x-1,y-1)=(-1)^3+(-2)^3+(2+1)^3+(2+2)^3+……..+5^3+(6+1)^3+(6+2)^3+……..+(z-1)^3 -(5+1)^3-(5+2)^3-……..-(x-1)^3 – (2+1)^3-(2+2)^3-……..-(y-1)^3

    Suppose:
    z^n=x^n+y^n.
    Special case:
    z^3=x^3+y^3.
    So:
    [z(z+1)/2]^2 – [x(x+1)/2]^2 – [y(y+1)/2]^2=[z(z-1)/2]^2 – [x(x-1)/2]^2 – [y(y-1)/2]^2.
    So:
    f(z,x,y)=f(z-1,x-1,y-1)=g(z,x,y)=g(z-1,x-1,y-1)!!!!!!!!

    And more.
    z^3=[z(z+1)/2]^2 – [a(a-1)/2]^2-(a+1)^3-(a+2)^3-……..-(z-1)^3 – a^3..
    So;
    with infinity value of integer (a) have infinity forms of z^3 that equal series cube integers.
    How found two integers x and y that satisfies infinity equations.

    ADIEU.

  2. Dear everybody.

    (N).

    With two any integers (a) > ( b)::
    a^3=[a(a+1)/2]^2 – [b(b+1)/2]^2 – [(b+1)^3+(b+2)^3+……..+(a-1)^3].
    So:::
    x^3=[x(x+1)/2]^2 – [3(4)/2]^2 – [4^3+5^3+……..+(x-1)^3].
    y^3=[y(y+1)/2]^2 – [4(5)/2]^2 – [5^3+6^3+……..+(y-1)^3].
    z^3=[z(z+1))/2]^2 – [5(6)/2]^2 – [6^3+7^3+……..+(z-1)^3].
    So
    4^3=[x(x+1)/2]^2 – [3(4)/2]^2 – [5^3+6^3+……..+(x-1)^3]. – x^3
    5^3=[y(y+1)/2]^2 – [4(5)/2]^2 – [6^3+7^3+……..+(y-1)^3]. – y^3.
    6^3=[z(z+1))/2]^2 – [5(6)/2]^2 – [7^3+8^3+……..+(z-1)^3]. – z^3.

    Because:
    6^3 – 3^3 – 4^3 – 5^3=0.
    So:
    [z(z+1))/2]^2 – [5(6)/2]^2 – [7^3+8^3+……..+(z-1)^3]. – z^3. – 3^3 – [x(x+1)/2]^2 + [3(4)/2]^2 |+ [5^3+6^3+……..+(x-1)^3]+x^3 – [y(y+1)/2]^2 + [4(5)/2]^2 + [6^3+7^3+……..+(y-1)^3]. +y^3=0.
    So;
    z^3 – x^3 – y^3.=[z(z+1))/2]^2 – [5(6)/2]^2 – [7^3+8^3+……..+(z-1)^3. – 3^3 – [x(x+1)/2]^2 + [3(4)/2]^2 |+ [5^3+6^3+……..+(x-1)^3]- [y(y+1)/2]^2 + [4(5)/2]^2 + [6^3+7^3+……..+(y-1)^3]..

    We had had this equation for any trio integers (z,x and y) , if had an extra equation z^3 – x^3 – y^3=0.
    Then:
    z,x and y can not be integers.

    ADIEU.

  3. (N+).
    Two formulas
    z^3=[z(z+1)/2]^2 – [z(z-1)/2]^2
    And :
    x^3+y^3=[y(y+1)/2]^2 – [x(x-1)/2]^2 – [(x+1)^3+(x+2)^3+……..+(x+a-1)^3+(x+a)^3+(x+a+1)^3+……..+(y-1)^3].

    Because:
    (x+a)^3=[(x+a)(x+a+1)/2]^2 – [(x+a)(x+a-1)/2]^2.
    So:
    x^3+y^3=[y(y+1)/2]^2 – [x(x-1)/2]^2 – [(x+a)(x+a+1)/2]^2 + [(x+a)(x+a-1)/2]^2 – [(x+1)^3+(x+2)^3+……..+(x+a-1)^3+(x+a+1)^3+……..+(y-1)^3]

    Suppose:
    z^3=x^3+y^3.
    So:
    [z(z+1)/2]^2 – [z(z-1)/2]^2=[y(y+1)/2]^2 – [x(x-1)/2]^2 – [(x+a)(x+a+1)/2]^2 + [(x+a)(x+a-1)/2]^2 – [(x+1)^3+(x+2)^3+……..+(x+a-1)^3+(x+a+1)^3+……..+(y-1)^3]

    Attention!
    This equation have four invariant square integers . The rest will change over integers (a).
    Corresponding to a value of (a) will have a corresponding equation.
    Impossible found integers to satisfy too many equations.

    Example:
    62^2 -32^2=25^2-14^2 – c^2+d^2 – [e^3+f^3+g^3+h^3+t^3+r^3]
    62^2 -32^2=25^2-14^2 – p^2+q^2 – [e^3+f^3+g^3+h^3+k^3+r^3]
    62^2 -32^2=25^2-14^2 – t^2+n^2 – [e^3+f^3+g^3+h^3+t^3+s^3]
    62^2 -32^2=25^2-14^2 – h^2+k^2 – [v^3+f^3+g^3+h^3+t^3+r^3]
    Too many equations.
    How satisfied them by only integers !!!!!!!!
    So:
    z^3=/x^3+y^3
    Similar.
    z^n=/x^n+y^n.
    ADIEU.

  4. Attention:
    Only (N+).
    Fermat’s last Theorem z ^ 3 = x ^ 3 + y ^ 3 is capable exists a solution if fully meet the following conditions:
    First step:
    (1+2+3+4+……..+a)^2+(1+2+3+4+……..+b)^2=v^2.
    In fact, using the computer, this equation has the ability to survive.
    Second step:
    (1+2+3+4+……..+a+1)^2+(1+2+3+4+……..+b+1)^2=s^2.
    Third step:
    v=1+2+3+4+……..+c.
    In fact, using the computer, this equation has the ability to survive.
    Fourth step:
    s=1+2+3+4+……..d.
    Fifth step:
    d=c+1
    If all five steps are satisfied.This equation is capable of existence.
    [z(z+1)/2]^2 – [z(z-1)/2]^2=[x(x+1)/2]^2- [x(x-1)/2]^2+[y(y+1)/2]^2 – [y(y-1)/2]^2.
    Because:
    z^3=[z(z+1)/2]^2 – [z(z-1)/2]^2.
    Mean this equation is capable of existence.
    z^3=x^3+y^3.
    However, too hard to satify all five equations in same time..

    And more:
    Attention about series of number:
    1,3,6,10,15,21,28,36,45……..
    Recognize:
    10 and 15 are two number consecutive which belong this string.
    Having:
    15^2 – 10^2=5^3.
    Or:
    z^3=[z(z+1)/2]^2 – [z(z-1)/2]^2.
    Impossible in same time exist both:
    [z(z+1)/2]^2=[x(x+1)/2]^2+[y(y+1)/2]^2
    And
    [z(z-1)/2]^2=[x(x-1)/2]^2+[y(y-1)/2]^2
    Attention:
    All numbers as z(z+1)/2 and x(x+1)/2 and y(y+1)/2 and z(z-1)/2 and x(x-1)/2 and y(y-1)/2 are belong this string and they are Pythagorean

    This is main proof:
    z^3=[z(z+1)/2]^2 – [z(z-1)/2]^2
    Define:
    x<x+a<y.
    x^3+y^3=[y(y+1)/2]^2 – [x(x-1)/2]^2 – [(x+1)^3+(x+2)^3+……..+(x+a-1)^3+(x+a)^3+(x+a+1)^3+……..+(y-1)^3]

    Suppose:
    z^3=x^3+y^3.
    Because also:
    (x+a)^3= [(x+a)(x+a+1)/2]^2 – [(x+a)(x+a-1)/2]^2.
    Therefore a system of equations is generated.
    [z(z+1)/2]^2 – [z(z-1)/2]^2=[y(y+1)/2]^2 – [x(x-1)/2]^2 – [(x+a)(x+a+1)/2]^2 + [(x+a)(x+a-1)/2]^2 – [(x+1)^3+(x+2)^3+……..+(x+a-1)^3+(x+a+1)^3+……..+(y-1)^3]

    [z(z+1)/2]^2 – [z(z-1)/2]^2=[y(y+1)/2]^2 – [x(x-1)/2]^2 – [(x+b)(x+b+1)/2]^2 + [(x+b)(x+b-1)/2]^2 – [(x+1)^3+(x+2)^3+……..+(x+b-1)^3+(x+b+1)^3+……..+(y-1)^3]

    [z(z+1)/2]^2 – [z(z-1)/2]^2=[y(y+1)/2]^2 – [x(x-1)/2]^2 – [(x+c)(x+c+1)/2]^2 + [(x+c)(x+c-1)/2]^2 – [(x+1)^3+(x+2)^3+……..+(x+c-1)^3+(x+c+1)^3+……..+(y-1)^3]

    [z(z+1)/2]^2 – [z(z-1)/2]^2=[y(y+1)/2]^2 – [x(x-1)/2]^2 – [(x+d)(x+d+1)/2]^2 + [(x+d)(x+d-1)/2]^2 – [(x+1)^3+(x+2)^3+……..+(x+d-1)^3+(x+d+1)^3+……..+(y-1)^3].

    ……..
    Can not count the number of equations

    And more;
    Using the two formulas by rotation affect each other:
    z^3=[z(z+1)/2]^2 – [z(z-1)/2]^2
    And
    [z(z+1)/2]^2=1^3+2^3+……..+z^3.
    This method makes root equation z^3=x^3+y^3 is structured as a Transfiguration equation which having unlimited formats as Robot Bumblebee Transformer.
    General case:
    z ^ n = x ^ n + y ^ n.
    Mean:
    z ^ (n-3) ​​* z ^ 3 = x ^ (n-3) ​​* x ^ 3 + y ^ (n-3) ​​* y^ 3. 
    You can structure this equation according to your own discretion when use z^3=[z(z+1)/2]^2 – [z(z-1)/2]^2 then use [z(z+1)/2]^2=1^3+2^3+……..+z^3 then continue by same to become your own format.
    ADIEU.

  5. Attention:
    Only (N+).
    Fermat’s last Theorem z ^ 3 = x ^ 3 + y ^ 3 is capable exists a solution if fully meet the following conditions:
    First step:
    (1+2+3+4+……..+a)^2+(1+2+3+4+……..+b)^2=v^2.
    In fact, using the computer, this equation has the ability to survive.
    Second step:
    (1+2+3+4+……..+a+1)^2+(1+2+3+4+……..+b+1)^2=s^2.
    Third step:
    v=1+2+3+4+……..+c.
    In fact, using the computer, this equation has the ability to survive.
    Fourth step:
    s=1+2+3+4+……..d.
    Fifth step:
    d=c+1
    If all five steps are satisfied.This equation is capable of existence.
    [z(z+1)/2]^2 – [z(z-1)/2]^2=[x(x+1)/2]^2- [x(x-1)/2]^2+[y(y+1)/2]^2 – [y(y-1)/2]^2.
    Because:
    z^3=[z(z+1)/2]^2 – [z(z-1)/2]^2.
    Mean this equation is capable of existence.
    z^3=x^3+y^3.
    However, too hard to satify all five equations in same time..

    And more:
    Attention about series of number:
    1,3,6,10,15,21,28,36,45……..
    Recognize:
    10 and 15 are two number consecutive which belong this string.
    Having:
    15^2 – 10^2=5^3.
    Or:
    z^3=[z(z+1)/2]^2 – [z(z-1)/2]^2.
    Impossible in same time exist both:
    [z(z+1)/2]^2=[x(x+1)/2]^2+[y(y+1)/2]^2
    And
    [z(z-1)/2]^2=[x(x-1)/2]^2+[y(y-1)/2]^2
    Attention:
    All numbers as z(z+1)/2 and x(x+1)/2 and y(y+1)/2 and z(z-1)/2 and x(x-1)/2 and y(y-1)/2 are belong this string and they are Pythagorean

    This is main proof:
    z^3=[z(z+1)/2]^2 – [z(z-1)/2]^2
    Define:
    x<x+a<y.
    x^3+y^3=[y(y+1)/2]^2 – [x(x-1)/2]^2 – [(x+1)^3+(x+2)^3+……..+(x+a-1)^3+(x+a)^3+(x+a+1)^3+……..+(y-1)^3]

    Suppose:
    z^3=x^3+y^3.
    Because also:
    (x+a)^3= [(x+a)(x+a+1)/2]^2 – [(x+a)(x+a-1)/2]^2.
    Therefore a system of equations is generated.
    [z(z+1)/2]^2 – [z(z-1)/2]^2=[y(y+1)/2]^2 – [x(x-1)/2]^2 – [(x+a)(x+a+1)/2]^2 + [(x+a)(x+a-1)/2]^2 – [(x+1)^3+(x+2)^3+……..+(x+a-1)^3+(x+a+1)^3+……..+(y-1)^3]

    [z(z+1)/2]^2 – [z(z-1)/2]^2=[y(y+1)/2]^2 – [x(x-1)/2]^2 – [(x+b)(x+b+1)/2]^2 + [(x+b)(x+b-1)/2]^2 – [(x+1)^3+(x+2)^3+……..+(x+b-1)^3+(x+b+1)^3+……..+(y-1)^3]

    [z(z+1)/2]^2 – [z(z-1)/2]^2=[y(y+1)/2]^2 – [x(x-1)/2]^2 – [(x+c)(x+c+1)/2]^2 + [(x+c)(x+c-1)/2]^2 – [(x+1)^3+(x+2)^3+……..+(x+c-1)^3+(x+c+1)^3+……..+(y-1)^3]

    [z(z+1)/2]^2 – [z(z-1)/2]^2=[y(y+1)/2]^2 – [x(x-1)/2]^2 – [(x+d)(x+d+1)/2]^2 + [(x+d)(x+d-1)/2]^2 – [(x+1)^3+(x+2)^3+……..+(x+d-1)^3+(x+d+1)^3+……..+(y-1)^3].

    ……..
    Can not count the number of equations

    And more;
    Using the two formulas by rotation affect each other:
    z^3=[z(z+1)/2]^2 – [z(z-1)/2]^2
    And
    [z(z+1)/2]^2=1^3+2^3+……..+z^3.
    This method makes root equation z^3=x^3+y^3 is structured as a Transfiguration equation which having unlimited formats as Robot Bumblebee Transformer.
    Similar,this method is used for general case:
    z ^ n = x ^ n + y ^ n.
    Mean:
    z ^ (n-3) ​​* z ^ 3 = x ^ (n-3) ​​* x ^ 3 + y ^ (n-3) ​​* y^ 3. 
    You can structure this equation according to your own discretion when use z^3=[z(z+1)/2]^2 – [z(z-1)/2]^2 then use [z(z+1)/2]^2=1^3+2^3+……..+z^3 then continue by same to become your own format.
    ADIEU.

  6. Fermat’s last Theorem is the transformer bumblebee Robot.
    Original equation:
    z^n=x^n+y^n.
    Mean:
    z^(n-3)*z^3=x^(n-3)*x^3+y^(n-3)*y^3.
    Using the formula z^3=[z(z+1)/2]^2 – [z(z-1)/2]^2 to convert z^3 become the exponent 2. Then using the formula [z(z+1)/2]^2=1^3+2^3+……..+z^3 to convert the exponent 2 become the exponent 3.
    Repeated several times with the same method.
    The transformer bumblebee Robot was created according your own structure.

    Simplest format about Transformer Bumblebee Robot.
    Using two formulas:
    z^3=[z(z+1)/2]^2 – [z(z-1)/2]^2
    And define x<x+a<y.
    x^3+y^3=[y(y+1)/2]^2 – [x(x-1)/2]^2 – [(x+1)^3+(x+2)^3+……..+(x+a-1)^3+(x+a)^3+(x+a+1)^3+……..+(y-1)^3]
    Because:
    (x+a)^3= [(x+a)(x+a+1)/2]^2 – [(x+a)(x+a-1)/2]^2
    So also:
    x^3+y^3=[y(y+1)/2]^2 – [x(x-1)/2]^2 – [(x+a)(x+a+1)/2]^2 + [(x+a)(x+a-1)/2]^2 – [(x+1)^3+(x+2)^3+……..+(x+a-1)^3+(x+a+1)^3+……..+(y-1)^3]

    Original equation:
    z^3=x^3+y^3.
    According to above method, the transformer bumblebee Robot system was created:
    [z(z+1)/2]^2 – [z(z-1)/2]^2=[y(y+1)/2]^2 – [x(x-1)/2]^2 – [(x+a)(x+a+1)/2]^2 + [(x+a)(x+a-1)/2]^2 – [(x+1)^3+(x+2)^3+……..+(x+a-1)^3+(x+a+1)^3+……..+(y-1)^3]

    [z(z+1)/2]^2 – [z(z-1)/2]^2=[y(y+1)/2]^2 – [x(x-1)/2]^2 – [(x+b)(x+b+1)/2]^2 + [(x+b)(x+b-1)/2]^2 – [(x+1)^3+(x+2)^3+……..+(x+b-1)^3+(x+b+1)^3+……..+(y-1)^3]

    [z(z+1)/2]^2 – [z(z-1)/2]^2=[y(y+1)/2]^2 – [x(x-1)/2]^2 – [(x+c)(x+c+1)/2]^2 + [(x+c)(x+c-1)/2]^2 – [(x+1)^3+(x+2)^3+……..+(x+c-1)^3+(x+c+1)^3+……..+(y-1)^3]

    [z(z+1)/2]^2 – [z(z-1)/2]^2=[y(y+1)/2]^2 – [x(x-1)/2]^2 – [(x+d)(x+d+1)/2]^2 + [(x+d)(x+d-1)/2]^2 – [(x+1)^3+(x+2)^3+……..+(x+d-1)^3+(x+d+1)^3+……..+(y-1)^3].

    ……..
    Flood robot on the planet Orion galaxy looking for the integers to eat.
    Certainly not enough the integer for the large number of transformer bumblebee Robot

    ADIEU.

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