(a) For which n > 2 is there a set of n consecutive positive integers such that the largest number in the set is a divisor of the least common multiple of the remaining n - 1 numbers?
(b) For which n > 2 is there exactly one set having this property?
Solution
(a) n = 3 is not possible. For suppose x was the largest number in the set. Then x cannot be divisible by 3 or any larger prime, so it must be a power of 2. But it cannot be a power of 2, because 2m - 1 is odd and 2m - 2 is not a positive integer divisible by 2m.
For k ≥ 2, the set 2k-1, 2k , ... , 4k-2 gives n = 2k. For k ≥ 3, so does the set 2k-5, 2k-4, ... , 4k-6. For k ≥ 2, the set 2k-2, 2k-3, ... , 4k-2 gives n = 2k+1. For k ≥ 4 so does the set 2k-6,2k-5, ... , 4k-6. So we have at least one set for every n ≥ 4, which answers (a).
(b) We also have at least two sets for every n ≥ 4 except possibly n = 4, 5, 7. For 5 we may take as a second set: 8, 9, 10, 11, 12, and for 7 we may take 6, 7, 8, 9 ,10, 11, 12. That leaves n = 4. Suppose x is the largest number in a set with n =4. x cannot be divisible by 5 or any larger prime, because x-1, x-2, x-3 will not be. Moreover, x cannot be divisible by 4, because then x-1 and x-3 will be odd, and x-2 only divisible by 2 (not 4). Similarly, it cannot be divisible by 9. So the only possibilities are 1, 2, 3, 6. But we also require x ≥ 4, which eliminates the first three. So the only solution for n = 4 is the one we have already found: 3, 4, 5, 6.
Solutions are also available in Murray S Klamkin, International Mathematical Olympiads 1978-1985, MAA 1986, and in István Reiman, International Mathematical Olympiad 1959-1999, ISBN 189-8855-48-X.
© John Scholes
jscholes@kalva.demon.co.uk
14 Oct 1998