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


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

$k n o w i n g t h a t 7^{3 k + 1} \equiv 7 [m o d 9]; k \in N .$ $s h o w t h a t 2005^{2005} \equiv 7 [m o d 9]$
	Answered by JDamian last updated on 27/Dec/20

	$2005 \equiv 7 m o d 9$ $2005 = 3 \times 668 + 1$ $2005^{2005} m o d 9 = 7^{3 \times 668 + 1} m o d 9$
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