2

I noticed that for the binary number $k=(1,10,11)$, $3^k$ begins with the digits of k. That relationship immediately breaks down, but it doesn't seem hard to find a power of 3 beginning similar to any binary number. E.g., $3^2$ starts with 100, $3^4$ starts with 101, etc.

Is this actually the case? Can every binary number be found at the start of some power of 3?

Joe Slater
  • 440
  • 2
  • 10
  • I’m unfamiliar with notating binary numbers as ordered triples. Could you explain that to me? – gen-ℤ ready to perish Dec 27 '17 at 22:47
  • @ChaseRyanTaylor I believe they were just writing down the binary representations of 1, 2, 3, and then making an observation about something that holds for these representations. – Mark Schultz-Wu Dec 27 '17 at 22:48
  • @Mark Ah, I see. Perhaps a better way way to phrase this would be, “. . . for a binary number $k\in{1,10,11}$, one has that $3^k$ begins with the digits of $k$.” (Obviously the “one has that” part is unnecessary, but I dislike separating mathematical expressions with only punctuation.) – gen-ℤ ready to perish Dec 27 '17 at 22:50
  • Would it be clearer if I amended it to say something like "The binary representation of $3^k$ begins with the same digits as the binary representation of k when k=1, k=2, or k=3"? – Joe Slater Dec 29 '17 at 02:39

1 Answers1

4

Yes. For each finite binary string starting with $1$ there is an interval $[a,b) \subset [0,1]$ such that $3^n$ begins with that string iff the fractional part of $\log_2(3^n)$ is in that interval. But because $\log_2(3)$ is irrational, the fractional parts of $\log_2(3^n) = n \log_2(3)$ are dense in $[0,1]$.

Robert Israel
  • 448,999