\( n \equiv 1 \pmod3, n \equiv 4 \pmod5 \): Solve \( n = 5k + 4 \). Then \( 5k + 4 \equiv 1 \pmod3 \Rightarrow 2k + 1 \equiv 1 \Rightarrow 2k \equiv 0 \Rightarrow k \equiv 0 \pmod3 \Rightarrow k = 3m \Rightarrow n = 15m + 4 \). So smallest is \( n = 4 \). - Groen Casting