1. (BEL) ${ }^{\mathrm{IMO} 4}$ (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?
1. Assume that the set $\{a-n+1, a-n+2, \ldots, a\}$ of $n$ consecutive numbers satisfies the condition $a \mid \operatorname{lcm}[a-n+1, \ldots, a-1]$. Let $a=p_{1}^{\alpha_{1}} p_{2}^{\alpha_{2}} \ldots p_{r}^{\alpha_{r}}$ be the canonic representation of $a$, where $p_{1} < p_{2} < \ldots < p_{r}$ are prime numbers and $\alpha_{1}, \alpha_{2}, \ldots, \alpha_{r} > 0$. Then for each $j=1,2, \ldots, r$, there exists $m, m=1,2, \ldots, n-1$, such that $p_{j}^{\alpha_{j}} \mid a-m$, i.e., such that $p_{j}^{\alpha_{j}} \mid m$. Thus $p_{j}^{\alpha_{j}} \leq n-1$. If $r=1$, then $a=p_{1}^{\alpha_{1}} \leq n-1$, which is impossible. Therefore $r \geq 2$. But then there must exist two distinct prime numbers less than $n$; hence $n \geq 4$. For $n=4$, we must have $p_{1}^{\alpha_{1}}, p_{2}^{\alpha_{2}} \leq 3$, which leads to $p_{1}=2, p_{2}=3$, $\alpha_{1}=\alpha_{2}=1$. Therefore $a=6$, and $\{3,4,5,6\}$ is a unique set satisfying the condition of the problem. For every $n \geq 5$ 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 $n \geq 6$. Let $r, s, t$ be natural numbers such that $2^{r} \leq n-1<2^{r+1}$, $3^{s} \leq n-1<3^{s+1}, 5^{t} \leq n-1<5^{t+1}$. Taking $a=2^{r} \cdot 3^{s}$ and $a=2^{r} \cdot 5^{t}$ we obtain two distinct sets with the required property. Thus the answers are (a) $n \geq 4$ and (b) $n=4$.