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 ? 
 A: 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.
A: 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.
A: 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.
A: 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
