HMMT February Team Round, 2019

Solution:
Answer: $1,2,3,4$

First, note that clearly there are no composite positive integers less than $2!$, and no $3$ consecutive composite positive integers less than $3!$. The only composite integers less than $4!$ are
$$4,6,8,9,10,12,14,15,16,18,20,21,22$$
and it is easy to see that there are no $4$ consecutive composite positive integers among them. Therefore, all $n \leq 4$ works.

Define $M=\operatorname{lcm}(1,2, \ldots, n+1)$. To see that there are no other such positive integers, we first show that for all $n \geq 5$, $n!>M$. Let $k=\left\lfloor\log_{2}(n+1)\right\rfloor$. Note that $v_{2}(M)=k$, while
$$v_{2}((n+1)!)=\sum_{i=1}^{k}\left\lfloor\frac{n+1}{2^{i}}\right\rfloor \geq \sum_{i=1}^{k}\left(\frac{n+1}{2^{i}}-1\right)=\left(n+1-\frac{n+1}{2^{k}}\right)-k \geq (n+1-2)-k=n-k-1.$$
This means that at least $(n-k-1)-k=n-2k-1$ powers of $2$ are lost when going from $(n+1)!$ to $M$. Since $M \mid (n+1)!$, when $n-2k-1 \geq k+1 \Longleftrightarrow n \geq 3k+2$, we have
$$M \leq \frac{(n+1)!}{2^{k+1}} \leq \frac{(n+1)!}{2(n+1)}<n!$$
as desired. Since $n \geq 2^{k}-1$, we can rule out all $k$ such that $2^{k} \geq 3k+3$, which happens when $k \geq 4$ or $n \geq 15$. Moreover, when $k=3$, we may also rule out all $n \geq 3k+2=11$.

We thus need only check values of $n$ between $5$ and $10$:

$n=5$: $n!=120$, $M=60$;

$n=6$: $n!=720$, $M=420$;

$n=7$: $n!=5040$, $M=840$;

$n \in \{8,9,10\}$: $n!\geq 40320$, $M \leq 27720$.

In all cases, $n!>M$, as desired.

To finish, note that $M-2, M-3, \ldots, M-(n+1)$ are all composite (divisible by $2,3, \ldots, n+1$ respectively), which gives the desired $n$ consecutive numbers. Therefore, all integers $n \geq 5$ do not satisfy the problem condition, and we are done.

Solution:
Here is a different way to show that constructions exist for $n \geq 5$. Note that when $n+1$ is not prime, the numbers $n!-2, n!-3, \ldots, n!-(n+1)$ are all composite (the first $n-1$ are clearly composite, the last one is composite because $n+1 \mid n!$ and $n!>2(n+1)$). Otherwise, if $n=p-1$ for prime $p \geq 7$, then the numbers $(n-1)!,(n-1)!-1,(n-1)!-2, \ldots,(n-1)!-(n-1)$ are all composite (the first one and the last $n-2$ are clearly composite since $(n-1)!>2(n-1)$, the second one is composite since $p \mid (p-2)!-1=(n-1)!-1$ by Wilson's theorem).