A property of every set of ten consecutive integers. In the following example of ten consecutive integers we can see that $119$ and $121$ are each coprime with the others:
$$114=2*3*19$$
$$115=5*23$$ $$116=2^2*29$$ $$117=3^2*13$$ $$118=2*59$$ $$119=7*17$$ $$120=2^3*3*5$$ $$121=11^2$$ $$122=2*61$$ $$123=3*41$$
This example illustrates a general property.
Prove that in every set of ten consecutive natural integers there is an integer (not necessarily unique) that is coprime with the other nine.
NOTE.- Someone of respectable reputation made a comment (which has been deleted) in stating that this beautiful result was false for which he quoted a post of StackExchange in which apparently had implied a denial of this property. I say this because I read that post slightly but did not keep the link. I appreciate much if someone makes me know this reference  I wanted to study.
 A: In such a set, there are exactly $5$ odd numbers. No more than two of them are multiple of three, and exactly one of them is multiple of five. There may be also a multiple of seven among the five odd numbers, but no more than one.
Therefore, at least one of the ten numbers $n$ is not a multiple of $2$, $3$, $5$ or $7$. The least prime factor of $n$ is greater than $10$ and thus, $n$ must be coprime with the other nine numbers of the set.
A: This addresses an ancillary question raised by ajotatxe in a comment beneath his answer. (It's too long to post as a comment itself.)  The question was whether there can be any string of consecutive numbers for which no entry is coprime to all the others.  I'll show there is such a string.
Consider a string of $20$ numbers $N, N+1, N+2,\ldots,N+19$, where $N$ satisfies the following congruences:
$$\begin{align}
N&\equiv0\mod(2\cdot3\cdot11\cdot13\cdot19)\\
N+7&\equiv0\mod5\\
N+5&\equiv0\mod7\\
N+1&\equiv0\mod17
\end{align}$$
Such an $N$ exists by the Chinese Remainder Theorem.  Clearly $N,N+2,N+4,\ldots,N+18$ all share a common factor of $2$.  As for the odd entries, $N+1$ shares a factor of $17$ with $N+18$; $N+3$, $N+9$, and $N+15$ share a factor of $3$; $N+5$ shares a factor $7$ with $N+12$; $N+7$ and $N+17$ share a factor of $5$; and $N+k$ for $k=11$, $13$, and $19$ share a factor of $k$ with $N$.
There may well be smaller examples; this construction was simply what came to mind.  It might, in fact, be of interest to find the first such string, or the shortest such string.  (It's not clear, to me at least, that the first such string is necessarily the shortest.)
