# Given any $10$ consecutive positive integers , does there exist one integer which is relatively prime to the product of the rest ?

Given any $10$ consecutive positive integers , does there exist one integer which is relatively prime to the product of the rest ?

• Yes, the question is : Is there an index $1 \le i \le 10$ such that $n+i$ is relatively prime to $\prod_{j=1, j\not=i}^{10} (n+j)$. Feb 9, 2015 at 14:50
• @user45878: yes , you have got it right
– user123733
Feb 9, 2015 at 14:53
• Every prime number larger than the largest integer in the sequence, would also be relatively prime to the product of that sequence. And of course, there are many other candidates (for example, every product of several prime numbers, all of which are larger than the largest integer in the sequence). Feb 9, 2015 at 15:32
• More generally, given $n$ consecutive integers, one of them is relatively prime to all the others, provided $n\lt17.$ There are counterexamples for every $n\ge17.$
– bof
Oct 20, 2015 at 6:40
• See this answer for literature on the claim in the prior comment. Mar 26, 2020 at 17:26

Notice that one integer $x$ being coprime to the product of the rest is equivalent to it being coprime to each other $y$ in the set, which is equivalent to it sharing no prime divisors. This gives immediately that:

$x$ cannot be divisible by $2$, $3$, or $5$, since for those primes, either $x+p$ or $x-p$ would have to be in the interval, given the interval has length $10$.

Clearly, we also have, for any prime $p$ equal to at least $11$ that if $x$ is in the interval $x+p$ and $x-p$ are not (and nor is any other multiple of $p$). Thus, if $x$ is also not divisible by $7$, it must be coprime to all the other integers in the desired interval.

However, any interval of length $10$ has at least one integer in it which is not divisible by $2$, $3$, $5$, or $7$. In particular, notice that there will be:

• $5$ integers divisible by $2$.
• At most $2$ odd integers divisible $3$.
• At most $1$ odd integer divisible by $5$.
• At most $1$ odd integer divisible by $7$.

Implying at least one integer is not divisible by any of those primes and hence satisfies the desired condition.

only common prime factors amongst $10$ con secutive 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 $5$

• one multiple of $7$ unaccounted for

for which makes $4$ 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.

Consider the totatives of $30: \{1,7,11,13,17,19,23,29\}$. Note that any consecutive set of ten numbers has at least two of these numbers relatively prime to 30. At most one of those will be divisible by $7$. The other one will not be divisible by any prime common to the other 9 numbers, since all its prime divisors are greater than $10$. Therefore is it relatively prime to the other numbers.

• $30$ is the largest $n$ such that all its totatives (coprimes) are primes (or $1$), i.e. such that if $1 < k < n\,$ and $\gcd(k,n)=1\,$ then $k$ is prime. $\ \$ Mar 26, 2020 at 17:20

I assume the integer has to be one of the 10. In which case I can think of cases in which there definitely is - if there is any prime in the sequence greater than our equal to 11 for example, or any integer which is a product of those primes (e.g. 143=13*11)

Whether this is true for all such sets of 10 consecutive numbers, I don't know

• This is more of a comment, as it doesn't answer the question. Feb 9, 2015 at 15:05