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, 2011

Number theory Difficulty 4.6 AIME Prove it South Africa

Show that for every natural number $n$ , there exists a natural number $k$ such that the number $kn$ starts with the digits 2011.

Solution

Suppose we wish to find a solution where $kn$ has $d+4$ digits. Then we must satisfy $2011 \cdot 10^d \le kn < 2012 \cdot 10^d$ , or equivalently
$k \in \left[ \frac{2011}{n} \cdot 10^d, \frac{2012}{n} \cdot 10^d \right).$
The size of this interval is $\frac{10^d}{n}$ . Clearly, we can choose $d$ sufficiently large such that the interval has size greater than 1, in which case it must contain an integer, and this integer will be a solution.

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Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.