Number theory · Prime factorization · Consecutive integers · Estimates
Let \(m > 1\) be a natural number. Prove that there is no sequence of \(2^{m}\) consecutive natural numbers all of which have exactly \(m\) prime factors, counted with multiplicity.
(For example, the number \(8000 = 2^{6} \cdot 5^{3}\) has \(6 + 3 = 9\) prime factors.)
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Serbian Municipal Competition 2020, high school grade I, category A, problem 4. Organized by the Mathematical Society of Serbia (DMS). Source