Many little divisors

A "solution" without Bertrand's Postulate:

Let $l(x)$ denote the lowest common multiple of all the positive integers $\le x$ . Although not rigorious for now, we should have a pretty good feeling that $l(\frac {\sqrt {n}}{2})\le n$ is only true for a finite number of $n$ . So far, we are quite certain that there exists a nonzero answer for this question.

We will show the following: $l(k)> 4(k+1)^2$ for all $k\ge 7$ . Note that this deduces $l(\frac {\sqrt {n}}{2})> n+1$ for $n\ge 256$ and should verify our speculation above.

Note that $l(k)=\prod_{i=1}^{\pi (k)} p_{i}^{\lfloor \log_{p_i}k\rfloor}$ where $p_i, \pi (x)$ denotes the $i$ th prime and the prime counting function respectively. we can bound this using $\lfloor x\rfloor >x-1$ and obtain $l(k)>\frac {k^{\pi (k)}}{\prod_{i=1}^{\pi (k)} p_{i}}>\frac {k^4}{210}>4(k+1)^2$ . The last inequality is true for $k\ge 30$ , thus we have proven our assertion for this range. One can check by hand the validity for $7\le k\le 29$ (a finite number of cases).

To solve the current problem, we take $k=6$ and $l(6)=60$ and our corresponding $n$ should lie inbetween $12^2, 14^2$ so $\boxed {180}$ is our answer.

Note: This is not a rigorous solution.

First, observe that $n = 180$ is a solution, since $\frac{\sqrt{n}}{2} \approx 6.7082$ , and $1, 2, 3, 4, 5, 6$ all divide $180$ . Thus, for any solution $n > 180$ , we must have $\frac{\sqrt{n}}{2} > 6$ and thus $1, 2, 3, 4, 5, 6$ must all divide $n$ . In other words, $60$ divides $n$ .

The next multiple of $60$ is $240$ . This doesn't satisfy the condition, since $\frac{\sqrt{240}}{2} \approx 7.7459$ and $7$ doesn't divide $240$ . Any larger solution must thus be divisible by $7$ as well; in other words, $420$ divides $n$ , so $n \ge 420$ .

Now, since $n \ge 420$ , we need $8$ to divide $n$ . This means we get something larger that divides $n$ , namely $840$ .

Now, since $n \ge 840$ , we need $11$ to divide $n$ . This means we get something larger that divides $n$ , namely $840 \cdot 11$ .

Now, since $n \ge 840 \cdot 11$ , we need $23$ to divide $n$ . This means we get something larger that divides $n$ , namely $840 \cdot 11 \cdot 23$ .

The pattern is clear. For $n \ge 840$ , we will always be able to find a prime number less than $\frac{\sqrt{n}}{2}$ but doesn't divide the current lower bound, thus multiplying the lower bound by the prime number. We can repeat this again. Since every time $\frac{\sqrt{n}}{2}$ at least doubles, by Bertrand's postulate we're guaranteed the existence of a new prime every step. This means there can be no greater $n$ satisfying this.

This "proves" that the largest $n$ is $\boxed{180}$ .