Exercise 4.5.9 (Among any ten consecutive integers, at least one is relatively prime to the the product of the others)

Proof. Consider a block $I=[[n,n+9]]$ of ten consecutive positive integers. If $n=1$, then $1\in J$ is relatively prime to the the product of the others. We can now assume that $n\ge 2$.

We prove first that there is an integer $s\in I$ which is not divisible by $2,3,5,7$.

The block $I$ contains exactly $5$ odd integers $m,m+2,m+4,m+6,m+8$. Consider the block $J=\{m,m+2,m+4,m+6,m+8\}\subset I$ of $5$ consecutive odd integers. Then $J$ contains at most $2$ multiples of $3$, at most one multiple of $5$ and at most one multiple of $7$. Therefore after elimination of these multiples, it remains at least one integer $s\in J$ which is not divisible by $3,5,7$. Since the elements of $J$ are odd,

Since $n\ge 2$, then $s\ge 2$, and every prime factor $p$ of $s$satisfies $p\ge 11$.

Every other element $t$ of the block $I$ is in the form $t=s\pm i$, where $i<10$. If $s$ and $t$ have some prime common divisor $p$, then $p\ge 11$. But $p\mid s$ and $p\mid t$, thus $p\mid t-s=\pm i$, so $p\mid i$, where $i<10$, thus $p<10$. Since $p\ge 11$, this is a contradiction. This proves that $s$ is relatively prime to every other element of the block $I$. Therefore $s$ is relatively prime to the product of the others elements of the block.

Among any ten consecutive integers, at least one is relatively prime to the the product of the others. □