Problem 4

(a) For which integers

n>2

does there exist a set of

n

consecutive positive integers such that the largest number in the set divides the least common multiple of the remaining

n-1

numbers? (b) For which integers

n>2

is there exactly one such set?

Step 3 of 6: First construction for

n>3

Previous step Next step

n>3: \quad k=(n-1)(n-2)=\mathrm{lcm}(n-1,n-2) \text{ works}

Detailed analysis

Take the largest element to be $k=(n-1)(n-2)$ , the product (equivalently lcm, since consecutive integers are coprime) of the two numbers just below it. Every prime power dividing $k$ divides $n-1$ or $n-2$ , both less than $n$ , so the bound from Step 1 is satisfied and the set of $n$ consecutive integers ending at $k$ works.