Đáp án: $2005^{2021}$ chia $21$ dư $14$
Giải thích các bước giải:
Ta có:
$2005\equiv 1(mod 3)$
$\to 2005^{2021}\equiv 1^{2021}(mod 3)$
$\to 2005^{2021}\equiv 1(mod 3)$
$\to 2005^{2021}=3k+1, k\in N$
Lại có:
$2005\equiv 3(mod 7)$
$\to 2005^{2021}\equiv 3^{2021}(mod 7)$
Mà $3^2\equiv 2(mod 7)$
$\to (3^2)^3\equiv 2^3(mod 7)$
$\to 3^6\equiv 8(mod 7)$
$\to 3^6\equiv 1(mod 7)$
$\to (3^6)^{336}\equiv 1^{336}(mod 7)$
$\to 3^{2016}\equiv 1(mod 7)$
$\to 3^{2016}\cdot 3^5\equiv 3^5(mod 7)$
$\to 3^{2021}\equiv 3^5\equiv 5(mod 7)$
$\to 2005^{2021}\equiv 5(mod 7)$
$\to 2005^{2021}=7q+5, q\in N$
$\to 3k+1=7q+5$
$\to 3k=7q+4$
$\to q$ chia $3$ dư $2$
$\to q=3p+2, p\in N$
$\to 2005^{2021}=7q+5=7(3p+2)=21p+14$
$\to 2005^{2021}$ chia $21$ dư $14$
