Stuttering calculator

Clearly 10 is being repeatedly by 3. For binary operations eg a*b,  a/b, this is the way I think it works.  I believe they have 2 registers holding the left hand value (displayed) and 1 holding the right hand. When = is hit  (left) operator (right) is evaluated and placed in the display (ie the  left register) and the right  is unchanged. Repeatedly hitting = continuously divides the display value by the right value in this case 3

yesbut how to explain the interesting numbes that pop up:


3.3333333333333333333333333333333

1.1111111111111111111111111111111

0.37037037037037037037037037037037

0.12345679012345679012345679012346

What's so special about that. Your last lint is missing an 8.

Seriously, you have found an interesting set of digits. I do not beleive there is any special reasoning behind them.
It is interesting to note that whenever you divide two intergers you will get an answer in which the digits will eventually repeat themselves.

its simpler, just divide 10 / 81 ;-)))

but you miss the other interesting strings

0,098765432098765432098765432098765... = 8 / 81 ;-)))

can be shown that the length of the repeating substring in <<1 / n>> is a divisor of the totient of <<n>>

in this case, <<9>> is a divisor of <<Phi( 81 ) = 81 - 27 = 54>>

some primes <<p>> have the longest repeating substring <<p - 1>>, being <<7>> the smallest one
<<1 / 7 = 0,14285714285714285714285714285714...>>

It's easy to see why these patterns occur if you try them using log division.

It's easy to see why these patterns occur if you try them using LONG division.

some primes <<p>> have the longest repeating substring <<p - 1>>, being <<7>> the smallest one
<<1 / 7 = 0,14285714285714285714285714285714...>>

IIRC:
What's even more interesting, TWO sevenths is: .285714285714....., FOUR sevents is .57142857....

Thanks for the points :-)