2. Find all natural numbers $a, b, c, d, e$ such that

(1) $a \neq 2$,

(2) $a<b<c<d<e$

(3) $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}+\frac{1}{e}=1$.
Solution. Let $a, b, c, d, e$ satisfy the conditions of the problem; then $a=3$, because: $a \neq 2$ according to (1); $a \neq 1$, because $a=1$ implies $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}+\frac{1}{e}>1$; $a \geq 4$ implies

$$
\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}+\frac{1}{e} \leq \frac{1}{4}+\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8}<1$, according to (2), and $b \geq 5$ implies

$$
\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}+\frac{1}{e} \leq \frac{1}{3}+\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8}<1$; $c=6$ implies that $d=7$. But $\frac{1}{e}=\frac{1}{4}-\frac{1}{7}$ has no solution in $\mathbb{N}$. Therefore, $c=5$. Then

$$
\frac{1}{d}+\frac{1}{e}=1-\frac{1}{3}-\frac{1}{4}-\frac{1}{5}=\frac{13}{60}
$$

and $5<d<e$. If $d \geq 9$ then $\frac{1}{d}+\frac{1}{e} \leq \frac{1}{9}+\frac{1}{10}<\frac{13}{60}$, which together with $5<d$ implies that $d=6, d=7$ or $d=8$. If $d \geq 7$, then $\frac{1}{e}=\frac{13}{60}-\frac{1}{d}$ has no solution in $\mathbb{N}$.

Therefore, $d=6$. From $\frac{1}{e}=\frac{13}{60}-\frac{1}{6}=\frac{1}{20}$ we get $e=20$. Thus, the required numbers are: $3,4,5,6,20$.