Question 60
Let $S \subset \{1, \dots, n\}$ be a nonempty set, where $n$ is a positive integer. We denote by $s$ the greatest common divisor of the elements of the set $S$. We assume that $s \neq 1$ and let $d$ be its smallest divisor greater than $1$. Let $T \subset \{1, \dots, n\}$ be a set such that $S \subset T$ and $|T| \geq 1 + \lfloor \frac{n}{d} \rfloor$. Prove that the greatest common divisor of the elements in $T$ is $1$.
Let $n$ ($n \geq 1$) be a positive integer and $U = \{1, \dots, n\}$. Let $S$ be a nonempty subset of $U$ and let $d$ ($d \neq 1$) be the smallest common divisor of all elements of the set $S$. Find the smallest positive integer $k$ such that for any subset $T$ of $U$, consisting of $k$ elements, with $S \subset T$, the greatest common divisor of all elements of $T$ is equal to $1$.
Let $n$ ($n \geq 1$) be a positive integer and $U = \{1, \dots, n\}$. Let $S$ be a nonempty subset of $U$ and let $d$ ($d \neq 1$) be the smallest common divisor of all elements of the set $S$. Find the smallest positive integer $k$ such that for any subset $T$ of $U$, consisting of $k$ elements, with $S \subset T$, the greatest common divisor of all elements of $T$ is equal to $1$.