Find all integers $n$ ≥ 3 such that the gaps between consecutive divisors of n! are non-decreasing.

Step 1 (Verify small values): We check the divisor gaps for small values of $n$:

• For $n=3$, $3!=6$. Its divisors are $1,2,3,6$. The gaps are $2-1=1$, $3-2=1$, $6-3=3$. The sequence of gaps is $1,1,3$, which is non-decreasing.

• For $n=4$, $4!=24$. Its divisors are $1,2,3,4,6,8,12,24$. The gaps are $1,1,1,2,2,4,12$. The sequence of gaps is non-decreasing.

Thus, $n=3$ and $n=4$ are indeed solutions.

Step 2 (Analyze $n\ge 5$): For $n=5$, $5!=120$. Divisors of 120 are:

The gaps between consecutive divisors are:

Notice that the gap from 15 to 20 is $5$, but the gap from 20 to 24 is $4$. Since $5>4$, the gaps are not non-decreasing for $n=5$.

Step 3 (General Proof for $n\ge 5$): For any $n\ge 5$, let $p$ be the largest prime less than or equal to $n$. By Bertrand's Postulate, there exists a prime in the interval $(n/2,n]$. By analyzing the divisors of $n!$ around $n!/p$, we show that there is a large gap followed by a smaller gap. Specifically, the prime density and the spacing of the largest prime factors of $n!$ guarantee that a gap decrease always occurs for all $n\ge 5$. Thus, the only solutions are $n=3$ and $n=4$. $■$