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):

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

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

\(=\left(3^4\right)^8-\left(3^3\right)^{10}-\left(3^2\right)^{14}\)

\(=3^{32}-3^{30}-3^{28}\)

\(=3^{28}.\left(3^4-3^2-1\right)\)

\(=3^{28}.71_{ }\)

=> \(81^8-27^{10}-9^{14}\) chia hết cho 71

Chia hết cho 79 nhé.

a) 8⁷-2¹⁸

=(2³)⁷-2¹⁸

=2²¹-2¹⁸

=2¹⁸.(2³-1)

=2¹⁸. 7

=2¹⁷.2.7

=2¹⁷.14  chia het cho 14

81^7-27^9-9^13 
=(3^4)^7-(3^3)^9-(3^2)^13 
=3^28-3^27-3^26 
=(3^26.3^2)-(3^26.3^1)-(3^26.1) 
=3^26.(9-3-1) 
=3^22.(3^4.5) 
=3^22.405 chia het cho 405 
=> 81^7-27^9-9^13 chia het cho 405

a.

8⁷ - 2¹⁸ = (2³)⁷ - 2¹⁸ = 2²¹ - 2¹⁸ = 2¹⁷ x (2⁴ - 2) = 2¹⁷ x (16 - 2) = 2¹⁷ x 14

Vậy 8⁷ - 2¹⁸ chia hết cho 14.

b.

81⁷ - 27⁹ - 9¹³ = (3⁴)⁷ - (3³)⁹ - (3²)¹³ = 3²⁸ - 3²⁷ - 3²⁶ = 3²² x (3⁶ - 3⁵ - 3⁴) = 3²² x 405

Vậy 81⁷ - 27⁹ - 9¹³ chia hết cho 405

a) 10⁶ - 5⁷

= 2⁶ . 5⁶ - 5⁷

= 5⁶ . (2⁶ - 5)

= 5⁶ . (64 - 5)

= 5⁶ . 59 chia hết cho 59

=> đpcm

b) 81⁷ - 27⁹ - 9¹³

= (3⁴)⁷ - (3³)⁹ - (3²)¹³

= 3²⁸ - 3²⁷ - 3²⁶

= 3²⁶ .(3² - 3 - 1)

= 3²⁶ . (9 - 3 - 1)

= 3²⁴ . 3² . 5

= 3²⁴ . 9 . 5

= 3²⁴ . 45 chia hết cho 45

=> đpcm

c) 8⁷ - 2¹⁸

= (2³)⁷ - 2¹⁸

= 2²¹ - 2¹⁸

= 2¹⁸ . (2³ - 1)

= 2¹⁸ (8 - 1)

= 2¹⁷ . 2 . 7

= 2¹⁷ . 14 chia hết cho 14

=> đpcm

d) 10⁹ + 10⁸ + 10⁷

= 10⁷ . (10² + 10 + 1)

= 5⁷ . 2⁷ . (100 + 10 + 1)

= 5⁷ . 2⁶ . 2 . 111

= 5⁷ . 2⁶ . 222 chia hết cho 222

=> đpcm