Question

[Tú Văn Võ]

Chứng tỏ rằng 81^8-27^10-9^14 chia hết cho 71

Answer 1

[Aki Tsuki]

Có:

+) \(81^4\equiv60\left(mod71\right)\)

\(\left(81^4\right)^2\equiv60^2\equiv50\left(mod71\right)\) (1)

+) \(27^5\equiv20\left(mod71\right)\)

\(\left(27^5\right)^2\equiv20^2\equiv45\left(mod71\right)\) (2)

+) \(9^7\equiv54\left(mod71\right)\)

\(\left(9^7\right)^2\equiv54^2\equiv5\left(mod71\right)\) (3)

Từ (1), (2), (3):

\(\Rightarrow81^8-27^{10}-9^{14}\equiv50-45-5\equiv0\left(mod71\right)\)

=> \(81^8-27^{10}-9^{14}⋮71\left(đpcm\right)\)