`color{#D0D0D0}{O}color{#C8C8C8}{m}color{#C0C0C0}{i}color{#B8B8B8}{s}color{#B0B0B0}{e}`

Cách 1:

Ta có:

`3^1` chia `7` dư `3`

`3^2` chia `7` dư `2`

`3^3` chia `7` dư `6`

`3^4` chia `7` dư `4`

`3^5` chia `7` dư `5`

`3^6` chia `7` dư `1`

`3^7` chia `7` dư `3` 

...

`=>` Chu kỳ số dư là `6`

Ta lại có: `120` chia `6` dư 0

Vậy `3^120` chia `7` dư `1`

``

Ta có:

`2^1` chia `11` dư `2`

`2^2` chia `11` dư `4`

`2^3` chia `11` dư `8`

`2^4` chia `11` dư `5`

`2^5` chia `11` dư `10`

`2^6` chia `11` dư `9`

`2^7` chia `11` dư `7`

`2^8` chia `11` dư `3`

`2^9` chia `11` dư `6`

`2^10` chia `11` dư `1`

`2^11` chia `11` dư `2`

...

`=>` Chu kỳ số dư là `10`

Ta lại có: `2024` chia `10` dư `4`

Vậy `2^2024` chia `11` dư `5` 

`-` `-` `-` `-` `-` `-` 

Cách 2:

Ta có:

$3^1 \equiv 3 \pmod{7}$

$3^2 \equiv 2 \pmod{7}$

$3^3 \equiv 6 \pmod{7}$

$3^6 \equiv (3^3)^2 \equiv 6^2 \equiv 36 \equiv 1 \pmod{7}$

$3^{120} = (3^6)^{20} \equiv 1^{20} \equiv 1 \pmod{7}$

Vậy $3^{120}$ chia `7` dư `1`

``

Ta có:

$2^1 \equiv 2 \pmod{11}$

$2^2 \equiv 4 \pmod{11}$

$2^3 \equiv 8 \pmod{11}$

$2^4 \equiv 16 \equiv 5 \pmod{11}$

$2^5 \equiv 32 \equiv 10 \equiv -1 \pmod{11}$

$2^{10} \equiv (2^5)^2 \equiv (-1)^2 \equiv 1 \pmod{11}$

$2^{2024} = 2^{2020} \cdot 2^4 = (2^{10})^{202} \cdot 2^4 \equiv 1^{202} \cdot 2^4 \equiv 1 \cdot 16 \equiv 16 \equiv 5 \pmod{11}$

Vậy $2^{2024}$ chia `11` dư `5`