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Consider the sequence $1,2,4,8,...,a^n = 2^n,...$ of all the powers of $2$. Prove that, given any digit $i ∈ {1,...,9}$, there exist infinitely many values of $n$ for which $a^n$ starts with that digit.

It only comes to my mind to use Poincaré's Recurrence Theorem, but I can not do that. Can anyone give a tip

Robert Shore
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Ilovemath
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  • "there exist infinitely many values of $n$ for which an starts with that digit." I assume you mean there are infinitely many values of $a_n = 2^n$ that start with the digit. If you meant $n$ starts with that digit there is no reason to introduce the $a_n = 2^n$ values at all. – fleablood Mar 14 '19 at 19:04
  • Use the fact that the leading digit of $x > 1$ is determined by ${\log_{10} x}$, where ${x} = x - \lfloor{x}\rfloor$ is the fractional part of $x$. – anomaly Mar 14 '19 at 19:15

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Brute force approach:

Write all terms in scientific notation, i.e. as $a\cdot 10^k$ for integer $k$ and $1\leq a<10$. Assume for contradiction that only finitely many terms of the sequence begin with $3$. In other words that only finitely many terms have $a\in[3,4)$ in our scientific notation.

That also means that only finitely many terms can have $a\in[6,8)$, as all such terms require the term before them to begin with $3$. Continuing with the same idea, we get that only finitely many terms have $a\in[1.2, 1.6)$. And so on.

It doesn't take too many of these intervals together to cover all of $[1,10)$, meaning there can only be finitely many terms, which is clearly a contradiction. The same argument holds for any other starting digit.

Arthur
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