Lời giải:
Theo định lý Fermat nhỏ ta có:
\(6^{11-1}\equiv 1\pmod {11}\)
\(\Leftrightarrow 6^{10}\equiv 1\pmod {11}\Rightarrow (6^{10})^{59}\equiv 1\pmod {11}\)
\(\Rightarrow 6^{590}\equiv 1\pmod {11}\Rightarrow 6^{592}\equiv 6^2\equiv 36\pmod {11}\)
\(\Rightarrow 6^{592}+8\equiv 36+8\equiv 44\equiv 0\pmod {11}\)
Hay: \(6^{592}+8\vdots 11\) (đpcm)