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Question Number 127132 by mathocean1 last updated on 27/Dec/20

knowing that 7^(3k+1) ≡7[mod 9] ;k ∈ N.  show that 2005^(2005) ≡7[mod 9]

$${knowing}\:{that}\:\mathrm{7}^{\mathrm{3}{k}+\mathrm{1}} \equiv\mathrm{7}\left[{mod}\:\mathrm{9}\right]\:;{k}\:\in\:\mathbb{N}. \\ $$$${show}\:{that}\:\mathrm{2005}^{\mathrm{2005}} \equiv\mathrm{7}\left[{mod}\:\mathrm{9}\right] \\ $$

Answered by JDamian last updated on 27/Dec/20

2005≡7 mod 9  2005=3×668+1  2005^(2005) mod 9=7^(3×668+1) mod 9

$$\mathrm{2005}\equiv\mathrm{7}\:{mod}\:\mathrm{9} \\ $$$$\mathrm{2005}=\mathrm{3}×\mathrm{668}+\mathrm{1} \\ $$$$\mathrm{2005}^{\mathrm{2005}} {mod}\:\mathrm{9}=\mathrm{7}^{\mathrm{3}×\mathrm{668}+\mathrm{1}} {mod}\:\mathrm{9} \\ $$

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