knowing-that-7-3k-1-7-mod-9-k-N-show-that-2005-2005-7-mod-9-

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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