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2-k-1-3-k-2-0-mod-k-5-min-k-k-N-




Question Number 146130 by Huy last updated on 11/Jul/21
  (2^k +1)(3^k +2)≡0(mod k+5)   min k=?   (k∈N)
$$\:\:\left(\mathrm{2}^{\mathrm{k}} +\mathrm{1}\right)\left(\mathrm{3}^{\mathrm{k}} +\mathrm{2}\right)\equiv\mathrm{0}\left(\mathrm{mod}\:\mathrm{k}+\mathrm{5}\right) \\ $$$$\:\mathrm{min}\:\mathrm{k}=?\:\:\:\left(\mathrm{k}\in\mathbb{N}\right) \\ $$

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