1. Let $a, b, c, d, e, f$ be a permutation of $1, 2, 3, 4, 5, 6$. If $2|\overline{a b}, 3| \overline{a b c}, 4|\overline{a b c d}, 5| \overline{a b c d e}$, $61 \overline{a b c d e f}$, find $\overline{a b c d e f}$.
(1982, Beijing Mathematical Competition (Grade 8))
(It is easy to know that $b$, $d$, $f$ are a permutation of $2$, $4$, $6$, and thus $a$, $c$, $e$ are a permutation of $1$, $3$, $5$. From $5 \mid \overline{a b c d e}$ we get $e=5$, so, $(a, c)=(1,3)$ or $(3,1)$. By enumeration, we get $\overline{a b c d e f}=123654$ or 321654.)