Question

CMR: 2222⁵⁵⁵⁵ + 5555²²²² chia hết cho 7 (dùng đồng dư mod)

Answer 1

Ta có:

\(2222\equiv-4\left(mod7\right)\Rightarrow2222^{5555}\equiv\left(-4\right)^{5555}\left(mod7\right)\left(1\right)\)

\(5555\equiv4\left(mod7\right)\Rightarrow5555^{2222}\equiv4^{2222}\left(mod7\right)\left(2\right)\)

Từ (1) và (2) \(\Rightarrow2222^{5555}+5555^{2222}\equiv\left(-4\right)^{5555}+4^{2222}\left(mod7\right)\)

Mà (-4)⁵⁵⁵⁵ + 4²²²² = -4²²²².(4³³³³ - 1) = -4²²²².[(4³)¹¹¹¹ - 1] = -4²²²².(64¹¹¹¹ - 1)

Lại có: \(64\equiv1\left(mod7\right)\Rightarrow64^{1111}\equiv1\left(mod7\right)\)

\(\Rightarrow64^{1111}-1\equiv1-1\left(mod7\right)\) hay \(64^{1111}-1⋮7\)

\(\Rightarrow-4^{2222}.\left(64^{1111}-1\right)⋮7\)

hay \(2222^{5555}+5555^{2222}⋮7\left(đpcm\right)\)