That's a fun game!: 1->1 10->10 11->(blank) 100->1 101->101 110->0 111->1 1000->10 1001->(blank) 1010->1010 1011->10 1100->(blank) 1101->01 1110->10 1111->(blank) One question that comes to mind is, if you randomly pick an integer between 2^(n-1) and 2^n-1, what's the expected number of uneliminated bits that remain, or equivalently, what is the expected number of bit-runs of odd length? (If you prefer, you can look at all bit strings of length n, whether they begin with a 1 or a 0; the answer is the same.) I get the sequence of answers 1, 1, 3/2, 3/2, ...; I'm guessing the pattern continues as 2, 2, 5/2, 5/2, etc. There are standard techniques for problems like this via recurrence relations, but I suspect there's a cute direct proof; does anyone see it? Jim Propp On Wednesday, May 25, 2016, David Wilson <davidwwilson@comcast.net <javascript:_e(%7B%7D,'cvml','davidwwilson@comcast.net');>> wrote:
To tell if a binary numeral is divisible by 3: Repeatedly remove adjacent pairs of matching digits (00 or 11) until remaining digits are alternating. If number of 1's in remaining number a multiple of 3, so was original number.
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