Ivan Shishkin, Rye (1878)

Problems/Number theoryReviewed

n2026n^{2026} commence par 2026

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Existe-t-il n∈N∗n \in \mathbb{N}^* tel que l’écriture en base 1010 de n2026n^{2026} commence par 20262026 ?

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Solution by ForkyEN

Discussions2 useful votes

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

Solution by Dattier

Discussions0 useful votes

  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}

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