24 May
2016
24 May
'16
11:53 a.m.
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.