From $n \equiv 5 \pmod6$, $n = 6k + 5$. Substitute into the first congruence: $6k + 5 \equiv 3 \pmod7 \Rightarrow 6k \equiv -2 \pmod7 \Rightarrow 6k \equiv 5 \pmod7$. Multiply both sides by the modular inverse of 6 mod 7 (which is 6, since $6 \times 6 = 36 \equiv 1 \pmod7$): $k \equiv 5 \times 6 \equiv 30 \equiv 2 \pmod7$. Thus, $k = 7m + 2$, and $n = 6(7m + 2) + 5 = 42m + 17$. The two-digit solutions are when $m = 0$: $n = 17$ and $m = 1$: $n = 59$. The smallest is $\boxed17$. - go-checkin.com