Problem (Algebra):
In a sequence of $n$ consecutive positive integers, where $n>1$, an element of the sequence is said to be cromulent if it is relatively prime to all other numbers in the sequence. Every element of a sequence with $n=2$ is cromulent because any two consecutive integers are relatively prime to each other.
Find the maximum and minimum possible number of cromulent elements in a sequence of $n$ consecutive positive integers with $n=7$.

Solution:
The minimum number is 1 and the maximum number is 3 . One example of a sequence of length 7 with one cromulent element is $4,5,6,7,8,9,10$, where 7 is the cromulent element. To show that it is not possible for such a sequence to have zero cromulent elements, consider two cases. If the sequence begins with an even number, then it contains three odd numbers. At most one of these is divisible by 3 , and at most one is divisible by 5 , so one of the odd numbers must be divisible by neither 3 nor 5 . This odd number differs by at most 6 from each other element of the sequence, so the only prime factors it can share with another element of the sequence are 2, 3, and 5 . Because it is divisible by none of these primes, it follows that the odd number in question is cromulent. Similarly, if the sequence begins with an odd number, then it contains four odd numbers; at most two of these are divisible by 3 , and at most one is divisible by 5 , so again, one odd number in the sequence must be divisible by neither 3 nor 5 . By the same argument, this element is cromulent.

One example of a sequence of length 7 with three cromulent elements is $1,2,3,4,5,6$, 7 , where 1,5 , and 7 are all cromulent.

To prove that a sequence of length 7 cannot have four cromulent elements, consider that the cromulent elements would all have to be odd. At least one of these four odd elements must be a multiple of 3 . Because one of the even elements must also be a multiple of 3 , it is thus not possible for all four odd elements to be cromulent.

Final answer: 1,3