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Quote: "It can be either 0 or 1, based on the way it is written."
The Base Number Can Only be 0 for Infinite/Recurring 0.999... Because if the Base Number was 1 ? it would No longer be Continuous!!
The Numbers After the . are What are Running Infinitely!
Hi Stangerzv 1/0 = 1 Because No Value has been Divided into the Start Value of 1
I am sure the Universe is Bigger than 1 ... But 1 what! ? 1 x the Largest Space we Can Calculate! But then we are Back to Estimated Math!
Quote: "I think so. But I don't understand the logic by which 1,2,3,4,5,6 + 2,3,4,5,6,7 is worse 1,2,3,4,5,6 + 7,8,9,10,11,12.
Reply All the Groups of Any 6 Numbers can be Good!? But How many times do we Repeat Each Number...
Within Any of our Best from 21 Numbered Balls! from Our Groups of 21 Combinations of 6 ?
Quote: "A frog starts hopping at point A. His first hop is 1 metre. His second is 1/2 metre. His third is 1/4 metre. And so on, each hop half the size of the one before.
He is aiming to reach point B which is 2 metres from A. Does he get there?
Reply...Not if it get's Run over by a Car Etc.
Let's make this Clearer!...
Pi = Has a Base Number of 3 Followed by Infinite .(Random Numbers)
Infinite/Recurring 0.999... Has a Base Number of 0 Followed by Infinite .(9's)
So the Differences are the Base Numbers & the Following Infinite Numbers!
Pi has a Brother... Infinite/Recurring 0.999...
What is the Difference!? Pi = ...(Random) Infinite/Recurring 0.999... = ...(9)
Have to Correct You Bob! There can only be a finite number of stars!... If the Whole Universe has been Mapped!?
Which won't Happen in Your life time or Mine!
This should Help any Doubters!...And is a Very Good example Why Estimated Math is BAD Math!
And the Reason Why Infinite/Recurring 0.999... = 1 ? Is Estimated Math!
http://blogs.discovermagazine.com/80beats/2010/12/01/the-estimated-number-of-stars-in-the-universe-just-tripled/
This Sounds like a Real Problem!
But!...0.9999....(recurring) is like the Math Universe! for 0.9999....(recurring) to Equal 1
Would be the same as Gathering All the Stars in the Universe ?? and then putting them in a Box and Calling the Box 1
To Help Solve the Problem Below I have Laid out the Numbers Etc. Using a Grid!
So We have 21 x 6 = 126 Spaces/Boxes to fill with Our Best 21 Numbered Balls.
If we make a Grid like a Chess Board Numbered... Starting from the Bottom Left Corner Columns A - F and the Rows upwards 1 - 21
We now have Reference points which may Help to make a Formula!?
Because we want to use All the Numbers at least Once then the first Column A1 to A21 Should contain the first 21 Balls 1 to 21
This is a Good Starting Point! Now what is the Best way to fill the Other 105 Spaces/Boxes with any of our Best 21 Balls ??
Looks like this has Stumped Everyone on this Math Forum! for the Moment ?
This Below might help Someone trying to Solve this Problem!??
http://lottery.merseyworld.com/cgi-bin/lottery?days=2&Machine=Z&Ballset=0&order=1&show=1&year=-1&display=NoTables
Hi
I am Trying to Find the Best Most Efficient way to Select Groups of 6 Balls from What May be the Best 21 Balls (Using Stats Etc.) from the 49 Balls in a Draw! Even though the 49 Balls are all the Same...Over a Period of Time All should come up at some time!
With a Maximum Selection of 21 Groups of 6 Balls (Using the Best 21 Balls) Which equals 21 Tickets at 1 Pound or 1 Euro Etc. Per Ticket.
So a Bad Selection Method would be... let's say we Chose the Numbers 4 & 12 in Every Group of 6 which would be the same Pair Chose 21 times? and the other 4 Numbers in every Group of 6 different! if possible.
The reason it's Bad is that if the Numbers 4 & 12 are not Drawn then we would lose on All 21 Tickets!
Hello Everyone!
Here is a Nice little Problem "Is there a Best Any 6 Numbers from 21 From a Lotto Using the Normal 49 Balls" ?
There are Many ways to Pick Any 6 from 21 below is a Couple of Examples!
(1) Use 3 Different Any 6 from 7 Balls = Total of 21 Combinations Using 21 Balls
(2) Use 7 Groups of 3 Balls Then Pick Any 2 Groups of 3 Balls = 21 Combinations of 2 Using 21 Balls
What I am after is a Formula that Shows the Most Efficient Method!?