6
$\begingroup$

enter image description here

Attempt:

The sequence $2^k$ $(k = 1, 2, . . .)$ is infinite, while the set of residual classes modulo $1000$ is finite, so there are two different integers $n < m$ such that $2^n ≡ 2^m \pmod {1000}$.

  • 6
    All I can tell you --- Is that this was most likely a contest problem from 1987.. – 2017-01-23
  • 1
    All I can tell you --- is that this is solvable by a computer program but I'm not sure how this would be done by hand. (These pieces of advice are so useful, right? :P) – 2017-01-23
  • 1
    All I can tell you --- Is that $775$, $ 2911$, and $5047$ are solutions for $n$. – 2017-01-23
  • 2
    @EhsanM.Kermani: I think $\bmod {1000}$ gives the right three digits assuming the ones place is on the right. – 2017-01-23
  • 1
    Possible duplicate of [Is $2^k = 2013...$ for some $k$?](http://math.stackexchange.com/questions/544214/is-2k-2013-for-some-k) – 2017-01-23
  • 1
    Possible duplicate of [Starting digits of $2^n$.](http://math.stackexchange.com/questions/13131/starting-digits-of-2n) – 2017-01-23

1 Answers 1

4

This is certainly not the intended solution if this was originally a contest problem, but here's one argument. Let $\{x\}$ denote the fractional part of $x$. The leading $4$ digits of $2^n$ are $1987$ iff \begin{align*} \log_{10^4} 1987 \leq \{n \log_{10^4} 2\} < \log_{10^4} 1988. \end{align*} The map $x \to x + \log_{10^4} 2$ is uniquely ergodic on $\mathbb{R}/\mathbb{Z}$, so such an $n$ exists; in fact, the set of such $n$ has density $\log_{10^4}(1988/1987)$.

  • 1
    +1 for using the phrase "uniquely ergotic" to answer a contest number theory problem! – 2017-01-23
  • 1
    @StellaBiderman: Noted, but this sort of thing is a classic problem from ergodic number theory, and the theory gives the result without any further work. It's one thing to try to work with as simple tools as are necessary, but ergodic theory really is the perfect tool for this sort of problem. – 2017-01-23
  • 0
    oh I 100% agree. I wasn't being facetious or anything. – 2017-01-23
  • 0
    @StellaBiderman: No problem, I'm not going to complain about getting a +1. :) – 2017-01-23
  • 1
    this is common in contest problems, there is an entire section on this technique in putnam and beyond. – 2017-01-23
  • 1
    @JorgeFernándezHidalgo: I think the result that $x + n\alpha$ is equidistributed on $\mathbb{R}/\mathbb{Z}$ for $\alpha$ irrational and $x$ arbitrary is a well-known classical result, at least. – 2017-01-23