Clearly the only common prime factors amongst 10 consecutive positive integers will be 2,3,5,7. 5 of them will be divisible by 2, at least one of which must also be divisible by 3 and at least one of which must also be divisible by 5, and, if two of the numbers are divisible by 7, one of them will be even. This leaves two multiples of 3, one multiple of 5, and one multiple of 7 unaccounted for, enough factors to dish out to 4 more of our integers. But this still leaves one integer not divisible by 2,3,5,7, and therefore co-prime with all other integers of the set, and so coprime with their product.

Can we replace $ 10$ by $ n$? I have found that this fails for $ 25$, although I am not sure of the other values. Can anyone give anything?

Can you post your example for 25? Maybe that gives us a clue.

I have read some where that this result is true for n up to 16 consecutive integers. I'll look up the source and post soon.

peter wrote: Can you post your example for 25? Maybe that gives us a clue. Well my example is the natural numbers from $ 1$ to $ 25$. MayankM wrote: I have read some where that this result is true for n up to 16 consecutive integers. I'll look up the source and post soon. The source of the problem said IHH pp.211 and so I took that out and I found what MayankM had siad to be true.

manjil wrote: peter wrote: Can you post your example for 25? Maybe that gives us a clue. Well my example is the natural numbers from $ 1$ to $ 25$. Really? I think $ 23$ is relatively prime to the product of the others though...

MayankM wrote: I have read some where that this result is true for n up to 16 consecutive integers. I'll look up the source and post soon. it's true. 16 can be done and 17 can't. the counterexample for $ n=17$ are consecutive numbers of which the first equals $ 27830$. (in fact the first can be $ 27830+30030k$ for a natural $ k$, mybe there exists a smaller example that i missed but the point is that there are infinitely many) u can proove it for $ n=16$ basicaly same as ilthigore has done for $ n=10$ above only a bit more complicated. $ n=17$ let $ a_1,...a_{17}$ be these consecutive numbers. if we set $ a_1$ to be devisable by $ 2,5,11$, $ a_2$ devisable by $ 3$, $ a_3$ by $ 7$ and $ a_4$ by $ 13$. it's easy to check thet none of them is than relatively prime to all others. we can calculate $ 27830$ with china reminder theorem now. what is left to do is to show that if there is a counter wxample for $ n$ than we can also find a counter example for $ n+1$. im working on this now.

As mentioned above, the statement is true when 'ten' is replaced by any positive integer $m \le 16$. It is false, however, if one replaces 'ten' by any integer $m > 16$. References for this fact: In a paper that you can find here: https://www.ias.ac.in/article/fulltext/seca/011/01/0006-0012, Pillai proved that it is true for $m \le 16$ and it is false for all $m$ with $17 \le m \le 430$. In a paper that you can find here: https://www.ams.org/journals/bull/1941-47-04/S0002-9904-1941-07455-0/S0002-9904-1941-07455-0.pdf, Brauer proves that it is false for all $m \ge 300$, thereby finishing the problem.

Note that if $p>10$ is a prime then it can divide at most $1$ number among these $10$ consecutive numbers. The rest is just casework considering the divisors $2,3,5,7$.