Question

Tìm số dư \(22^{22}+55^{55};3^{1993}\) cho 7

Answer 1

Ta có:

\(22\equiv1\left(mod7\right)\Leftrightarrow22^{22}\equiv1\left(mod7\right)\)(1)

Mặt khác \(55\equiv6\left(mod7\right)\Leftrightarrow55^{55}\equiv6^{55}\left(mod7\right)\)

Mà \(6^2\equiv1\left(mod7\right)\)(2)

tách: \(6^{55}=6^{2.27+1}=\left(6^2\right)^{27}.6\equiv1^{27}.6=6\)(từ (2) ) (3)

Từ (1) và (3) suy ra \(22^{22}+55^{55}\) chia 7 dư 0

2) Ta có:

\(3^6\Leftrightarrow1\left(mod7\right)\)

tách: \(3^{1993}=3^{6.332+1}=\left(3^6\right)^{332}.3\equiv1^{332}.3=3\)(mod 7)
Vậy \(3^{1993}\) chia 7 dư 3