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1200-year-old problem

Topic closed. 16 replies. Last post 10 years ago by JADELottery.

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JADELottery's avatar - YingYangYong 01.PNG
The Quantum Master
West Concord, MN
United States
Member #21
December 7, 2001
3684 Posts
Offline
Posted: December 16, 2006, 5:26 pm - IP Logged

Mathematics Forum:
The place to discuss the strong ties between lotteries and mathematics.

Excuse me, but I think I'm missing the strong tie between lotteries and mathematics in this thread.

This problem actually came up in a topic a while back -->  The combinatorics of a permutation

 

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Now, the tricky math part.

Using the values of MIN, MAX, SUM, and PRODUCT we can find the values of X_LOW and X_HIGH.

Taking the SUM of W, X, Y, and Z, if we subtract the values of MIN and MAX we are left with the sum of the two middle numbers X_LOW and X_HIGH.


      Sum of X_LOW and X_HIGH:    n = SUM - MIN - MAX


Taking the PRODUCT of W, X, Y, and Z, if we divide out the values of MIN and MAX we are left with the product of the two middle numbers X_LOW and X_HIGH.


    Product of X_LOW and X_HIGH:    m = PRODUCT / (MIN * MAX)


At this point, we still don't know the values of X_LOW and X_HIGH.

However, the values can be represented as lowercase x and y.

It does not matter which is X_LOW or X_HIGH.

We can represent the sum of X_LOW and X_HIGH as:  x + y

We can also represent the product of X_LOW and X_HIGH as:  xy


      Sum of X_LOW and X_HIGH:    n = x + y

    Product of X_LOW and X_HIGH:    m = xy


We now use these equations to solve for one of the variables, preferably x.

First, isolate y in the sum equation.


        n = x + y

      n - x = x + y - x

      n - x = y

        y = n - x


Next, substitute the value of y into the product equation.


        m = xy

        m = x( n - x )

        m = nx - xx

        m = nx - x2


You math geeks should see this is a Quadratic Equation.

Now, put the Quadratic Equation in Standard Form, ax2 + bx + c = 0.


    -( nx - x2 ) + m = nx - x2 - ( nx - x2 )

        x2 - nx + m = nx - x2 - nx + x2

        x2 - nx + m = 0


From the Quadratic Equation in Standard From, we see that a = 1, b = -n, and c = m.

Using the Quadratic Formula, we can solve for x.


                      -b ± √ b2 - 4ac 
    Quadratic Formula:    x =  ---------------
                          2a


                      -(-n) ± √ (-n)2 - 4(1)(m) 
                  x =  -------------------------
                              2(1)


                      n ± √ n2 - 4m 
                  x =  -------------
                          2



The values of x are the two roots of the Quadratic Equation, this occurs from the ± symbol.

The ± symbol can also determine the X_LOW and X_HIGH values.



                          n - √ n2 - 4m 
      The X_LOW value is:      x =  -------------
                              2
                          n + √ n2 - 4m 
    The X_HIGH value is:      x =  -------------
                              2



We can now substitute the values of n and m from the previous equations n = SUM - MIN - MAX and m = PRODUCT / (MIN * MAX).



                          (SUM - MIN - MAX) - √ (SUM - MIN - MAX)2 - 4(PRODUCT / (MIN * MAX)) 
      The X_LOW value is:      x =  -------------------------------------------------------------------
                                                2



                          (SUM - MIN - MAX) + √ (SUM - MIN - MAX)2 - 4(PRODUCT / (MIN * MAX)) 
    The X_HIGH value is:      x =  -------------------------------------------------------------------
                                                2
As you can see, if MIN is a zero, dividing the PRODUCT by zero leads to an error.

However, because we have added a one to the numbers, we avoid getting an error and can less the values of X_LOW and X_HIGH by one to get the PICK 4 numbers.


Continues
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Jehocifer

    JADELottery's avatar - YingYangYong 01.PNG
    The Quantum Master
    West Concord, MN
    United States
    Member #21
    December 7, 2001
    3684 Posts
    Offline
    Posted: December 16, 2006, 5:29 pm - IP Logged

    Also, I did a mthematical work out for the X_LOW and X_HIGH equations with 0 and treated 0 as if it were a variable and was able to get the right answer.

    Not by adding 1, but just working the equation like 0 is a variable.

    02 ¹ 0  and 3 * 0 ¹ 0

    02 ® 0 and 3 * 0 ® 0 are computational operations not mathematical operations.

    02 and 3 * 0 in an equation, the 0 is a variable and needs to be handled through mathematical equation solving operations not computational operations.

    It's also the same for ¥.

    Presented 'AS IS' and for Entertainment Purposes Only.
    Any gain or loss is your responsibility.
    Use at your own risk.

    Order is a Subset of Chaos
    Knowledge is Beyond Belief
    Wisdom is Not Censored
    Douglas Paul Smallish
    Jehocifer