I'm an engineer, not a mathematician or statistician, but I'm sure there's a logical explanation how this works:
milaadesign.com - website designer
milaadesign.com - website designer
Math/Statistics Wizards ---- Help!
- Thread starter cribbj
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Ian Anderson
Lifetime Supporter
It's magic
Ian Anderson
Lifetime Supporter
Actually not so
When you add the 2 digits together and subtract it from the number first thought of you will end up with a number divisible by 9
Put that simbol on 9, 18, 27 etc and it always works.
Example 10 1+0 = 1 10-1 = 9
Example 99 9+9 = 18 99-18= 81 again divisible by 9
Example 81 8+1 = 9 81 -9 = 72 again divisible by 9
Another one for you
take 1/7th = 0.142857
Take this number and multiply it by any number 1 - 10
you get the same digits - just in a different order
or the old number Pi (3.142857)
Square it and you get the acceleration due to gravity in meters per second
(Does that mean Pi on the moon would be a different number?)
Ian
When you add the 2 digits together and subtract it from the number first thought of you will end up with a number divisible by 9
Put that simbol on 9, 18, 27 etc and it always works.
Example 10 1+0 = 1 10-1 = 9
Example 99 9+9 = 18 99-18= 81 again divisible by 9
Example 81 8+1 = 9 81 -9 = 72 again divisible by 9
Another one for you
take 1/7th = 0.142857
Take this number and multiply it by any number 1 - 10
you get the same digits - just in a different order
or the old number Pi (3.142857)
Square it and you get the acceleration due to gravity in meters per second
(Does that mean Pi on the moon would be a different number?)
Ian
interesting, use algebra to prove Ians statement.
any two digit number can be expressed in tens and units,
say a x 10 + b x 1
so when the sum of a and b are subtracted you get
a x 10 + b - (a + b) = 9 x a
so QED any two digit number will return a factor of 9
All the numbers in the table are redundant except the factors of 9, which all use the same symbol.
Ian, pi^2 is = gravity, is it just a coinpidence?
Dave
any two digit number can be expressed in tens and units,
say a x 10 + b x 1
so when the sum of a and b are subtracted you get
a x 10 + b - (a + b) = 9 x a
so QED any two digit number will return a factor of 9
All the numbers in the table are redundant except the factors of 9, which all use the same symbol.
Ian, pi^2 is = gravity, is it just a coinpidence?
Dave
My favourite 'mathematical coincidence' I recall from 'A' level days:-
e^(i*pi) = -1
or e^(i*pi) + 1 = 0 as Euler wrote.
Of course it's not a coincedence at all and can easily be shown how this relationship occurs. It is quite hard to try & explain to a non-mathematical person why linking these fundamental constants in this way to give such a nice 'complex but simple' formula is at all interesting!
e^(i*pi) = -1
or e^(i*pi) + 1 = 0 as Euler wrote.
Of course it's not a coincedence at all and can easily be shown how this relationship occurs. It is quite hard to try & explain to a non-mathematical person why linking these fundamental constants in this way to give such a nice 'complex but simple' formula is at all interesting!
Thanks Ian, I worked it out algebraically to be a number divisible by 9, but couldn't sort out the smoke & mirrors on the webpage, until I noticed the symbols were identical for the series of numbers divisible by 9, dohhhh.
Each time you "play" it, the software shuffles the symbols to confuse things a bit, but it always puts identical symbols on the numbers divisible by 9.
I'm a little slow today.......
Each time you "play" it, the software shuffles the symbols to confuse things a bit, but it always puts identical symbols on the numbers divisible by 9.
I'm a little slow today.......
Slightly off-topic, but can anyone solve Y=X^X for X= f(Y)? So simple an equation, but I've never seen a solution. And the graph of the function for negative X is bizarre and real for even integers only.
John,
I had a go taking logs, but got stuck, then I googled it and came up with the following thread
Solving y=x^x for x Text - Physics Forums Library
I got to step 4, but lacked the knowledge to complete it
regards
Dave
I had a go taking logs, but got stuck, then I googled it and came up with the following thread
Solving y=x^x for x Text - Physics Forums Library
I got to step 4, but lacked the knowledge to complete it
regards
Dave
I think using the Lambert function, which is essentially the inverse of Y=Xe^X, is a bit circular. The Google solution X = exp[W[ln[y]]] seems more like a restatement of the problem, since W is a functional expression of both X and Y, but then I haven't done this stuff for a long time.
