Ivan Shishkin, Rye (1878)

Problems/Number theoryUnreviewed

n2026n^{2026} begins with 20262026

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Does there exist n∈N∗n \in \mathbb{N}^* such that the base-1010 representation of n2026n^{2026} begins with 20262026?

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Solution by DattierFR

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  1. Suffit de montrer que {m×2026ln⁡(3)+n×ln⁡(10):(m,n)∈N×Z}\{m\times 2026\ln(3)+ n \times \ln(10) : (m,n) \in \mathbb N \times \mathbb Z\} est dense dans R\mathbb R.

  2. Pour cela suffit de remarquer que 2026ln⁡(3)ln⁡(10)∉Q\dfrac{2026\ln(3)}{\ln(10)} \not \in \mathbb Q, car ∀(a,b)∈N∗,32026a≠10b\forall (a,b) \in \mathbb N^{*}, 3^{2026a}\neq 10^{b}

Solution by Forky

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The number X = n at power 2026 can be written as 10000x + 2026 = 2(5000x + 1013), with 5000x + 1013 as an odd number. X is a pefect square of n at power 1013. Then X cannot double of an odd number. No solution

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