Welcome Guest
Log In | Register )
You last visited December 8, 2016, 12:55 am
All times shown are
Eastern Time (GMT-5:00)

Are you good at Algebra?

Topic closed. 32 replies. Last post 4 years ago by Scientistman.

Page 3 of 3
PrintE-mailLink
Avatar
Kentucky
United States
Member #32652
February 14, 2006
7313 Posts
Offline
Posted: August 24, 2012, 2:56 pm - IP Logged

Ok hi this is scientistman I am the one who started the discussion. Thank you all for your input.

 

I will attempt to break it down somewhat.

I am only attempting to see if algebra might work for my formula.

I am not wanting to give my formual away so I hope if someone wins with it they might remember me.

I should have included more than x y and z so I will now.

We will use   v w x y and z if this is ok to do in algebra

I know (v) and  I know (w) and I know (x) I just dont know (y) and if I know (y) then I know (z)

(y) is somewhat unpredictible yet might be possibly attainable through algebra. Maybe?

If I can attain (y) I can attain the winning numbers in any lottery every single time.

 

 

So for the example

lets use 2 7 8 and 10 11 18

 

so it would look like this

 

02 07 08 is (v)

10 11 18 is (w)

_____________

= 8 4 10 (x)

so

10 - 2 = 8

11 - 7 = 4

18 - 8 = 10

I still dont know ( y )

so lets say (y) is 12 15 19, now 12 15 19 is not the answer its (y) and (y) is what you need to arrive at the answer, with (y) you can attain (z) or any winning lottery numbers anywhere in the world before it occurs.

but to arrive at (z) the answer! I need (y)

So if you know (y) you take (x) and (y) and subtract (v)

and this gives you  (z) The EXACT Lottery numbers Everytime! Without fail.

It's Right on the Money every time in any drawing anywhere in the world.

Without (y) it doesnt work, so you have to know (y) to arrive at (z)

so How would I arrive at (Y) ?

 Or am I asking the impossibly question? I'm not sure cause some people are brainiacs.

I just recall a story where a math teacher handed out the tests and had two mathimatical equations on the chalk board.

One student handed in his test and did one of the problems on thechaulk board, the teach freaked out what he saw because the 2 questions on the chaulk bard where not part of the test they were the 2 UNSOLVABLE math equations ofall time and the stuent figured out one of them thinking it was just part of the test.

I hope I am not confusing anyone. I took algebra and got straight A's but I think that was just because the math teacher felt sorry for me because of a family death in my family and the whole school including the teachers felt sad for me so I think the teacher passed me, so I forgot most of what I knew.

 

Im just wondering if (y) can be attained if you know that (v) has to be subtracted from it to arrive at (z) or do you need the actual answer (z) first to figure out the example ?

If the answer is IMPOSSIBLE without me giving you (z) then I understand I just wanted to throw it out there. Thanks

"10 - 2 = 8

11 - 7 = 4

18 - 8 = 10

I still dont know ( y )"

8 + y = z, 4 + y = z, 10 + y = z,

If z = 12, y must be 4, 8, or 2.

"(y) is somewhat unpredictible yet might be possibly attainable through algebra. Maybe?"

I'm assuming the value  of "x", "w", and "v is always known, but the value of "y" varies. Since "z" is the result, you can use past results to determine the past values of "y". If there is a consistent value of "y", use that number. If the value of "y" is a random number between 1 and 175 million, algebra can't help you.

    psykomo's avatar - animal shark.jpg

    United States
    Member #4877
    May 30, 2004
    5121 Posts
    Offline
    Posted: August 27, 2012, 1:15 pm - IP Logged

    Ok how would you find out what X is if you know Y and Z

    Whick came first:

    The Chicken R Chick Fil UR Lay !!??

    Y = Chicken

    X = Chick Fil UR Lay !!??

    LOL = JackPot

    DrumJack-in-the-BoxDrum


      United States
      Member #123632
      February 25, 2012
      156 Posts
      Offline
      Posted: September 4, 2012, 1:37 am - IP Logged

      "10 - 2 = 8

      11 - 7 = 4

      18 - 8 = 10

      I still dont know ( y )"

      8 + y = z, 4 + y = z, 10 + y = z,

      If z = 12, y must be 4, 8, or 2.

      "(y) is somewhat unpredictible yet might be possibly attainable through algebra. Maybe?"

      I'm assuming the value  of "x", "w", and "v is always known, but the value of "y" varies. Since "z" is the result, you can use past results to determine the past values of "y". If there is a consistent value of "y", use that number. If the value of "y" is a random number between 1 and 175 million, algebra can't help you.

      Yes Stack you are on the right track with me, "y" varies but "y" varies between 1 though 10 most of the time.

      So figuring out "y" is a bit like figuring out the next lotto numbers but at least I have it narrowed down to selecting the numbers between 1 through 10 and not selecting numbers between 1 through 49.

      So if I can figure out "y" which is between the digits 1 and 10 then I can figure out the next lotto winning numbers each and every time. If I can figure out "y" which is between 1 through 10 then I can use "y" and figure out "z"

       

      Z is the actual winning numbers

      Y are the digits between 1 and 10 that I use to arrive at Z

      I am now creating a database of past "y" results. And possibly can guess at "y" and if my guess is correct I can get z

      "y" is always known after the drawing but not before, but "y" is bewteen 1-10 not 1 through 49 so Ive narrowed it down to selecting numbers bewteen 1 through 10 but its still difficult unless I can find a way to zero in on the numbers between 1-10

      y consist of 6 numbers bewteen 1 and 10