MHB Co-prime Numbers in a Series of 10

[kaliprasad]

Show that in 10 consecutive numbers there is at least one number which is co-prime to other 9 numbers.

 

[Opalg]

[sp]In any set of ten consecutive integers, each of the numbers $0,1,\ldots,9$ will occur as the last digit of one of those integers. Select the four integers ending in $1$, $3$, $7$ and $9$. None of those will be divisible by $2$ or $5$. At most two of them will be divisible by $3$ (because consecutive odd multiples of $3$ differ by $6$), and at most one of them will be divisible by $7$ (because consecutive odd multiples of $7$ differ by $14$). So at least one of those four numbers, $N$ say, is not divisible by $3$ or $7$ (or by $2$ or $5$). Apart from $2$, $3$, $5$ and $7$, no other prime can be a factor of more than one integer in a consecutive run of ten. Therefore none of the prime factors of $N$ occurs in any of the other nine numbers, and so $N$ is coprime to all of them.[/sp]

 

[kaliprasad]

[sp]In any set of ten consecutive integers, each of the numbers $0,1,\ldots,9$ will occur as the last digit of one of those integers. Select the four integers ending in $1$, $3$, $7$ and $9$. None of those will be divisible by $2$ or $5$. At most two of them will be divisible by $3$ (because consecutive odd multiples of $3$ differ by $6$), and at most one of them will be divisible by $7$ (because consecutive odd multiples of $7$ differ by $14$). So at least one of those four numbers, $N$ say, is not divisible by $3$ or $7$ (or by $2$ or $5$). Apart from $2$, $3$, $5$ and $7$, no other prime can be a factor of more than one integer in a consecutive run of ten. Therefore none of the prime factors of $N$ occurs in any of the other nine numbers, and so $N$ is coprime to all of them.[/sp]

above answer is good and better than my answer which is as below

Spoiler

Out of 10 consecutive numbers 5 numbers are divisible by 2 and not more that 4 numbers are divisible by 3 out of which
maximum 2 numbers are odd and divisible by by 3 that makes 7, 2 numbers are divisible by 5 out of which is even so there are maximum one number is divisible by 5 and neither 2 nor 3 and that makes 8 and maximum 2 numbers are divisible by 7 out of which is only one is odd maximum one number is divisible by 7 and neither 2 nor 3 nor 5 and that makes maximum 9 numbers that are divisible by one of 2,3,5,7. so there is at least one number which is not divisible by 2,3,5 or 7 so the lowest prime factor of the same is 11 and it cannot divide any other number of the set. so this number is co-prime to rest 9.