\( n \equiv 2 \pmod3, n \equiv 1 \pmod5 \): Let \( n = 5k + 1 \). Then \( 5k + 1 \equiv 2 \pmod3 \Rightarrow 2k + 1 \equiv 2 \Rightarrow 2k \equiv 1 \Rightarrow k \equiv 2 \pmod3 \Rightarrow k = 3m + 2 \Rightarrow n = 5(3m+2)+1 = 15m + 11 \). Smallest is \( n = 11 \). - Crosslake