Tìm dư trong phép chia: a) 3^40 cho 82 b) 3^100 cho 7 c) 4362^4362 cho 11 Mọi người giúp mình với, mình đang cần gấp

`a)` `3^4=81≡-1 (mod 82)`

`=>(3^4)^10≡(-1)^{10} (mod 82)`

`=>3^{40}≡1(mod 82)`

Vậy `3^{40}` chia $82$ dư $1$

`b)`

`\qquad 3^2≡2(mod 7)`

`=>(3^2)^5≡2^5≡4(mod 7)`

`=>(3^{10})^2≡4^2≡2 (mod 7)`

`=>(3^{20})^5≡2^5(mod 7)`

`=>3^{100}≡4(mod7)`

Vậy `3^{100}` chia $7$ dư $4$

`c)` `4362≡6(mod 11)`

`=>4362^2≡6^2≡3(mod 11)`

`=>(4362^2)^5≡3^5≡1(mod 11)`

`=>(4362^{10})^{436}≡1^{436}≡1(mod 11)`

`=>4362^{4360}.4362^2≡1.3(mod 11)`

`=>4362^{4362}≡3(mod 11)`

Vậy `4362^{4362}` chia $11$ dư $3$