Câu hỏi của Hoàng Ngọc Tuyết Nung

Question

tìm dư trong phép chia $2004^{2004}$ cho 11

Kim Tuyến · Accepted Answer

Ta có: $2002\subset11=>2004-2\subset11$

$=>2004\equiv2\left(mod11\right)$

$=>2004^{2004}=2^{2004}\left(mod11\right)$ Mà $2^{10}\equiv1=>2004^{2004}=2^4.\left(2^{10}\right)^{200}\equiv24\equiv5\left(mod11\right)$

Vậy $2004^{2004}chia11$dư 5Câu trả lời: Ta có: $2002\subset11=>2004-2\subset11$

$=>2004\equiv2\left(mod11\right)$

$=>2004^{2004}=2^{2004}\left(mod11\right)$ Mà $2^{10}\equiv1=>2004^{2004}=2^4.\left(2^{10}\right)^{200}\equiv24\equiv5\left(mod11\right)$

Vậy $2004^{2004}chia11$dư 5