\(97^5\equiv37\left(mod51\right)\)
\(\left(97^5\right)^3\equiv37^3\equiv10\left(mod51\right)\)
\(\left(97^{15}\right)^4\equiv10^4\equiv4\left(mod51\right)\)
\(\left(97^{60}\right)^4\equiv4^4\equiv1\left(mod51\right)\)
\(\left(97^{240}\right)^{83}\equiv1^{83}\equiv1\left(mod51\right)\)
\(\Rightarrow97^{20021}\equiv97^{19920}\cdot97^{60}\cdot97^{15}\cdot97^{15}\cdot97^5\cdot97^5\cdot97\equiv1\cdot4\cdot10\cdot10\cdot37\cdot37\cdot46\equiv25189600\equiv37\left(mod51\right)\)
Vậy số dư trong phép chia trên là 37
Đúng 0
Bình luận (3)
