The "11's" trick should work fine in base 2 for detecting multiples of 3. Count the number of 1's in even positions and the numbers of 1's in odd positions; iff these differ by a multiple of 3, you have a multiple of 3. On Tue, May 24, 2016 at 10:52 AM, Dan Asimov <dasimov@earthlink.net> wrote:
Zak, good question: Is there a quick way to detect multiples of 3 expressed in base 2 ??? Let's see:
(2 + 1) (2^a + 2^b + ... + 2^c) =
2^(a+1) + 2^a + 2^(b+1) + 2^b + ... + 2^(c+1) + 2^c.
Hmm, if all these exponents are distinct, this pattern seems very easy to detect.
But what if some of the exponents are equal?
I bet there's a simple algorithm to untangle them.
But I may be wrong.
—Dan
On May 24, 2016, at 9:56 AM, Zak Seidov via math-fun < math-fun@mailman.xmission.com> wrote:
Also, in our native, base - 10, numbers divisible by 3 has sum of digits also divisible by 3, while base - 2 has nothing similar.
_______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
-- -- http://cube20.org/ -- [ <http://golly.sf.net/>Golly link suppressed; ask me why] --