\(3^{2^{1930}}=3^{2.2^{1929}}=9^{2^{1929}}\equiv2^{2^{1929}}\left(mod7\right)\)
Ta có : \(2^{1929}=2^{1928}.2=4^{964}.2\equiv2\left(mod3\right)\)
Do đó \(2^{1929}\) có dạng \(2^{1929}=3k+2\) \(\left(k\in N\right)\)
\(\Rightarrow2^{2^{1929}}=2^{3k+2}=8^k.4\equiv4\left(mod7\right)\)
Hay \(3^{2^{1930}}\equiv4\left(mod7\right)\)
Vậy \(3^{2^{1930}}\) chia \(7\) dư \(4\)