I am happy that my series 1,3,6,10,15,21 . . . has proven to be good. For the past few days I was thinking that it failed in the classic weights and scale problem ( the problem was what weights needed to measure any weight upto 100 kilos?). 1,3,6,10,15,21 . . . was great in rational number series. But it didn't do well in the classic scale and weights puzzle it seemed.
But 1,3,6 were good in upto 10 kilos category.
And 1,3,6,10 were good in upto 20 kilos.
But upto 100?
Then I thought about
1,3,6,
10,30,60,
100,300,600,
1000,3000,6000
And then it seemed to fit.
This is one solution to the classic scale and weights puzzle.
5 weights are needed in 3^n series.
6 weights in 1,3,6,10,30,60
7 weights in 2^n series.
This finished the puzzle.
Well the above gave me some satisfaction and I worked a continued fraction for pi.
I think 11 pi = 100÷(2+2^2÷(3+3^2÷(4+4^2÷(5+5^2 . . . . .
But 1,3,6 were good in upto 10 kilos category.
And 1,3,6,10 were good in upto 20 kilos.
But upto 100?
Then I thought about
1,3,6,
10,30,60,
100,300,600,
1000,3000,6000
And then it seemed to fit.
This is one solution to the classic scale and weights puzzle.
5 weights are needed in 3^n series.
6 weights in 1,3,6,10,30,60
7 weights in 2^n series.
This finished the puzzle.
Well the above gave me some satisfaction and I worked a continued fraction for pi.
I think 11 pi = 100÷(2+2^2÷(3+3^2÷(4+4^2÷(5+5^2 . . . . .
2 comments:
It is not 11 pi. 11 pi is correct only to 2 places. 110033/100000 is more better.
I wonder whether the result of the continued fraction is rational multiple of pi?
I don't think it is 11 pi. The result was 11.003309959 till 30 places.
Any the beautiful pi in continued fraction is
4÷(1+1^2÷(3+2^2÷(5+3^2÷(7+4^2 . . .
Post a Comment