[Algebra]
Problem: 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 $n=6$;

Solution: First we prove that every sequence of five consecutive positive integers contains a cromulent element.

Proof: Consider a sequence of five consecutive integers. Exactly one number in such a sequence will be a multiple of 5 , but that number could also be a multiple of 2 and hence share a common factor with at least one other number in the sequence. There are several cases to consider, namely whether the sequence starts with an even number or an odd number.

If the sequence starts with an even number, then the second and fourth numbers are both odd, and at least one of them is not a multiple of 3 and hence is relatively prime to all other numbers in the sequence because it is neither a multiple of 2 nor 3 and hence is at least 5 away from the nearest integer with a common factor. Thus the sequence contains a cromulent element.

If the sequence starts with an odd number, then again, it contains an odd number that is not a multiple of 3 and hence is relatively prime to all other numbers in the sequence, thus the sequence contains a cromulent element. In fact, it contains two such numbers if the first or last number is a multiple of 3 , and if the middle number is a multiple of 3 , then all three odd elements are cromulent.


The minimum number is 1 and the maximum number is 2 . One example of a sequence of length 6 with one cromulent element is $5,6,7,8,9$, 10, where 7 is the cromulent element. To show that it is not possible for


a sequence of six consecutive elements to have zero cromulent elements, consider two cases. If the sequence begins with an even number, that number is not cromulent, and one of the other five elements must be cromulent by the argument in the proof above. A similar argument establishes that one element must be cromulent if the sequence of length 6 begins with an odd number (and thus ends in an even number).

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

To prove that a sequence of length 6 cannot have three cromulent elements, consider that the cromulent elements would all have to be odd, and one of those three would be a multiple of 3 . Because one of the even elements must also be a multiple of 3 , it is not possible for all three odd elements to be cromulent.

Final answer: 1,2