Problem 3. Consider the set $A=\{n, n+1, n+2, \ldots, 2 n\}$, where $n \geq 4$ is a natural number. Determine the smallest value of $n$ for which $A$ contains five elements $a<b<c<d<e$ such that

$$
\frac{a}{c}=\frac{b}{d}=\frac{c}{e}
$$
Solution. Let $p, q \in \mathbb{N}^{*},(p, q)=1$ such that $\frac{a}{c}=\frac{b}{d}=\frac{c}{e}=\frac{p}{q}$. Clearly $p<q$. Since the numbers $a, b$ and $c$ are divisible by $p$, and the numbers $c, d$ and $e$ are divisible by $q$, it follows that there exists $m \in \mathbb{N}^{*}$ such that $c=m p q \ldots \ldots \ldots \ldots \ldots . \ldots$..............

Since $e-a \leq 2 n-n=n$, and $n$ must be minimal, the numbers $a, b$ and $c$ must be consecutive multiples of $p$, and $c, d$ and $e$ must be consecutive multiples of $q$. Therefore, $a=m p q-2 p$ and $e=m p q+2 q \ldots \ldots .1 \mathbf{p}$

Since $n \leq a<e \leq 2 n \leq 2 a$, it follows that $2 a \geq e$, i.e., $2 m p q-4 p \geq$ $m p q+2 q$, or $m p q \geq 4 p+2 q .(*)$

Since $\frac{c}{e}=\frac{m p \bar{q}}{m p q+2 q}=\frac{p}{q}$, we get $m(q-p)=2$, so $m \in\{1,2\} \ldots 1 \mathbf{p}$

If $m=1$, then $q-p=2$, so $q=p+2$. Substituting into (*) we get: $(p-2)^{2} \geq 8$, hence $p \geq 5$. For $p=5$ and $q=7$, we have $a=25, b=30$, $c=35, d=42, e=49$ and, since $n \leq a<e \leq 2 n$, it follows that $n=25 \ldots .2 \mathbf{p}$

If $m=2$, then $q-p=1$, so $q=p+1$. Substituting into $(*)$ we get: $(p-1)^{2} \geq 2$, hence $p \geq 3$. For $p=3$ and $q=4$, we have $a=18, b=21$,
$c=24, d=28, e=30$ and, since $n \leq a<e \leq 2 n$, it follows that $n \in\{16,17,18\}$. Therefore, $n_{\min }=16$