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2^n can begin with any sequence of digits

Source: Mathcenter Contest / Oly - Thai Forum 2009 R1 p5 https://artofproblemsolving.com/community/c3196914_mathcenter_contest

November 9, 2022
number theoryDigitspower of 2

Problem Statement

For n∈Nn\in\mathbb{N}, prove that 2n2^n can begin with any sequence of digits.
Hint: log⁡2\log 2 is irrational number.
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