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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 $2^n ≡ 2^m \pmod {1000}$.

Watson
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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 $\log_{10^4}(1988/1987)$.

anomaly
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    +1 for using the phrase "uniquely ergotic" to answer a contest number theory problem! – Stella Biderman Jan 23 '17 at 04:48
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    @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. – anomaly Jan 23 '17 at 04:50
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    oh I 100% agree. I wasn't being facetious or anything. – Stella Biderman Jan 23 '17 at 04:52
  • @StellaBiderman: No problem, I'm not going to complain about getting a +1. :) – anomaly Jan 23 '17 at 04:53
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    this is common in contest problems, there is an entire section on this technique in putnam and beyond. – Asinomás Jan 23 '17 at 05:07
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    @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. – anomaly Jan 23 '17 at 05:23