AoPS - Problem

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Tags: floor function, logarithms, Irrational numbers

Peter

25.05.2007 03:24

You are given three lists A, B, and C. List A contains the numbers of the form $10^{k}$ in base 10, with $k$ any integer greater than or equal to 1. Lists B and C contain the same numbers translated into base 2 and 5 respectively: \[\begin{array}{lll}A & B & C \\ 10 & 1010 & 20 \\ 100 & 1100100 & 400 \\ 1000 & 1111101000 & 13000 \\ \vdots & \vdots & \vdots \end{array}.\] Prove that for every integer $n > 1$, there is exactly one number in exactly one of the lists B or C that has exactly $n$ digits.

The number of digits of $n$ in base $b$ is given by $\left\lfloor \log_{b}(n) \right\rfloor +1 = \left\lfloor \frac{\log_{10}(n)}{\log_{10}(b)} \right\rfloor+1$. Especially our numbers $10^k$ have $\left\lfloor \frac{k}{\log_{10}(2)} \right\rfloor$ and $\left\lfloor \frac{k}{\log_{10}(5)} \right\rfloor$ digits. Now simply use Beatty's theorem ( http://www.problem-solving.be/pen/viewtopic.php?t=235 ) for $\alpha = \frac{1}{\log_{10}(2)}$ and $\beta = \frac{1}{\log_{10}(5)}$ (the condition is obviously fulfilled by $\log_{10}(2)+\log_{10}(5)=\log_{10}(10)=1$).