Number theory · Digit sums · Divisibility by 9 · Powers of two · Decimal digit count
Let \(k > 3\) and consider the number \(2^k\). Prove that no rearrangement of the decimal digits of \(2^k\) can produce the number \(2^n\) with \(n > k\).
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Serbian Republic Competition (Republicko takmicenje) 2003, high school grade I, category A, problem 2. Organized by the Mathematical Society of Serbia (DMS). Source