10 Consecutive Integers contain Coprime Integer

Theorem
Let $n \in \Z$ be an integer.

Let $S := \set {n, n + 1, n + 2, \ldots, n + 9}$ be the set of $10$ consecutive integers starting from $n$.

Then at least one element of $S$ is coprime to every other element of $S$.

Proof
Consider $2$ elements $a, b$ of $S$ which share a common divisor $d$.

Then $d \divides \size {a - b}$ and so $d < 10$.

Now from the Fundamental Theorem of Arithmetic, $d$ must have a prime factor which is strictly less than $10$.

So for $a$ and $b$ to have a common divisor, at least one such common divisor is in $\set {2, 3, 5, 7}$.

There are exactly $5$ elements of $S$ which have a common divisor of $2$.

There are either $3$ or $4$ elements of $S$ common divisor of $3$.

The case where there are $4$ happens when $n = 3 k$ and $n + 9 = 3 \paren {k + 3}$.

Of these $3$ or $4$, no more than $2$ are odd and so have not been accounted for.

There are exactly $2$ elements of $S$ which have $5$ as a common divisor.

One of those is even and so has been counted already.

There are at most $2$ elements of $S$ which have $7$ as a common divisor.

One of those is even and so has been counted already.

Thus we have a count of how many elements of $S$ which can possibly share a common divisor with another element of $S$:


 * $5$ are divisible by $2$
 * $2$ or fewer are divisible by $3$ and not $2$
 * $1$ at the most is divisible by $5$ and not by $2$ or $3$
 * $1$ at the most is divisible by $7$ and not by $2$, $3$ or $5$.

That makes a total of $9$ elements of $S$ which are divisible by $2$, $3$, $5$ or $7$.

Thus there exists (at least) $1$ element of $S$ which is not divisible by $2$, $3$, $5$ or $7$ and so cannot share a common divisor with another element of $S$.

Hence the result.