Câu hỏi của Sammie

Question

Tìm số dư trong phép chia: $2^{1999}$ cho 35

Aki Tsuki · Accepted Answer

Ta có: $2^{24}\equiv1\left(mod35\right)$

$\left(2^{24}\right)^{83}\equiv1^{83}\equiv1\left(mod35\right)$

$\Rightarrow2^{1992}\cdot2^7\equiv1\cdot23\equiv23\left(mod35\right)$

Vậy 21999 chia 35 dư 23Câu trả lời: Ta có: $2^{24}\equiv1\left(mod35\right)$

$\left(2^{24}\right)^{83}\equiv1^{83}\equiv1\left(mod35\right)$

$\Rightarrow2^{1992}\cdot2^7\equiv1\cdot23\equiv23\left(mod35\right)$

Vậy 21999 chia 35 dư 23

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