1981 IMO Problems/Problem 4

Problem

(a) For which values of $\displaystyle n>2$ is there a set of $\displaystyle n$ consecutive positive integers such that the largest number in the set is a divisor of the least common multiple of the remaining $\displaystyle n-1$ numbers?

(b) For which values of $\displaystyle n>2$ is there exactly one set having the stated property?

Solution

Let $k = \prod p_i^{e_i}$ be the greatest element of the set, written in its prime factorization. Then $\displaystyle k$ divides the least common multiple of the other elements of the set if and only if the set has cardinality at least $\displaystyle \max \{ p_i^{e_i} \}$ , since for any of the $p_i^{e_i}$ , we must go down at least to $k-p_i^{e_i}$ to obtain another multiple of $p_i^{e_i}$ . In particular, there is no set of cardinality 3 satisfying our conditions, because each number greater than or equal to 3 must be divisible by a number that is greater than two and is a power of a prime.

For $\displaystyle n > 3$ , we may let $\displaystyle k = \mbox{lcm} (n-1, n-2) = (n-1)(n-2)$ , since all the $\displaystyle p_i^{e_i}$ must clearly be less than $\displaystyle n$ and this product must also be greater than $\displaystyle n$ if $\displaystyle n$ is at least 4. For $\displaystyle n > 4$ , we may also let $\displaystyle k = \mbox{lcm} (n-2, n-3) = (n-2)(n-3)$ , for the same reasons. However, for $\displaystyle n = 4$ , this does not work, and indeed no set works other than $\displaystyle \{ 3,4,5,6 \}$ . To prove this, we simply note that for any integer not equal to 6 and greater than 4 must have some power-of-a-prime factor greater than 3.