> Describe how to count up to 1023 on 10 fingers. :)
132 to you!
Am I right in thinking that that 132 suggests strange bases?
I used the quadraditic equation to figure out the bases -- there are
two, one positive and one negative:
For any equation
ax^2 + bx + c = 0
then
--------
-b +- \/ b^2 -4ac
x = -----------------
2a
or in Emacs Lisp
(/ (+ ( - b) (sqrt (- (expt b 2) (* 4 a c)))) (* 2 a))
Given:
x� + 3x + 2 = 1023
then
x� + 3x - 1021 = 0
and
a = 1
b = 3
c = -1021
so
(let ((a 1.0)
(b 3.0)
(c -1021.0))
(/ (+ (- b) (sqrt (- (expt b 2) (* 4 a c)))) (* 2 a)))
--> 30.488279103446626
or
(let ((a 1.0)
(b 3.0)
(c -1021.0))
(/ (- (- b) (sqrt (- (expt b 2) (* 4 a c)))) (* 2 a)))
--> -33.48827910344663
and checking;
(+ (* 30.488279103446626 30.488279103446626) (* 3 30.488279103446626))
--> 1021.0
(+ (* -33.48827910344663 -33.48827910344663) (* 3 -33.48827910344663))
--> 1021.0000000000003
I would guess that people who prefer natural numbers, but who also
want to avoid 2, 8, 10, 12, and 16 would choose either 24 or 60 for
their base. In this case 1023 would be, using decimal notation for
the values of each power:
in base 24
1 18 15 --> 1023
and using an alphabetical notation following the numbers 0 - 9, where
10 is A and 11 is B, the value is:
1IF
i.e,, from
1*(24^2) + 18*24 + 15
or, in Emacs Lisp
(+ (* 1 24 24) (* 18 24) 15) --> 1023
in base 60
17 3 --> 1023
i.e,, from
17*60 + 3
or, using another alphabetical notation following the numbers 0 - 9,
where 10 is A and 11 is B, and 36 is BA (i.e., 9 + 26 + 1), with a
space marking each power, the value is:
H 3
or, in Emacs Lisp
(+ (* 17 60) 3) --> 1023
(I did not bother with any equation for the last computations, but
simply divided by 60 or 24 or 24-squared to find the values.)
--
Robert J. Chassell Rattlesnake Enterprises
http://www.rattlesnake.com GnuPG Key ID: 004B4AC8
http://www.teak.cc [EMAIL PROTECTED]
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