IMO 1981 SL 1

Origin: BEL

Problem

(a) For which values of 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 values of n > 2 is there a unique set having the stated property?

Solution

Assume that the set {a−n+ 1, a−n+2, . . ., a} of n consecutive numbers satisfies the condition a | lcm[a −n + 1, . . . , a −1]. Let a = p\alpha1 1 p\alpha2 2 . . . p\alphar r be the canonic representation of a, where p1 < p2 < \cdot \cdot \cdot < pr are primes and \alpha1, \cdot \cdot \cdot , \alphar > 0. Then for each j = 1, 2, . . ., r, there exists m, m = 1, 2, . . . , n −1, such that p\alphaj j | a −m, i.e., such that p\alphaj j | m. Thus p\alphaj j \leq n−1. If r = 1, then a = p\alpha1 \leqn−1, which is impossible. Therefore r \geq2. But then there must exist two distinct prime numbers less than n; hence n \geq4. For n = 4, we must have p\alpha1 1 , p\alpha2 \leq3, which leads to p1 = 2, p2 = 3, \alpha1 = \alpha2 = 1. Therefore a = 6, and {3, 4, 5, 6} is a unique set satisfying the condition of the problem. For every n \geq5 there exist at least two such sets. In fact, for n = 5 we easily find two sets: {2, 3, 4, 5, 6} and {8, 9, 10, 11, 12}. Suppose that n \geq6. Let r, s, t be natural numbers such that 2r \leqn −1 < 2r+1, 3s \leqn −1 < 3s+1, 5t \leqn −1 < 5t+1. Taking a = 2r \cdot 3s and a = 2r \cdot 5t we obtain two distinct sets with the required property. Thus the answers are (a) n \geq4 and (b) n = 4.