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.

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.
