$24$ is the largest integer divisible by all integers less than its square root

Question

Show that $24$ is the largest integer divisible by all integers less than its square root.

This is what I have done :

Let $m$ be the greatest integer such that $m^2\leq n$, so $i\mid n$ for all i $\in \{1,2,\cdots m-1,m\}$ so lcm$(1,2,\cdots,m-1,m)\,\mid n $.

But how do i show that this is not possible when $n\geq 25$ or $m\geq 5$.

Answer 1

Suppose that $n$ is divisible by every integer up to $N:=\lfloor \sqrt n\rfloor$. Then, in particular, $n$ is divisible by $N$ and by $N-1$, and since $N$ and $N-1$ are co-prime, $n$ is in fact divisible by $N(N-1)$. Furthermore, it is easy to verify that $N(N-1)\ne n$, so that $n\ge 2N(N-1)$. It follows that $$ n > 2(\sqrt n-1)(\sqrt n-2) = 2n - 6\sqrt n + 4, $$ implying $6\sqrt n > n + 4$. It follows easily that $(\sqrt n-3)^2<5$; equivalently, $n<14+6\sqrt 5\approx 27.416$, and it remains to verify that 25, 26, and 27 do not work.

Answer 2

Here's a hint / observation:

The number $25$ is not divisible by $2, 3,$ or $4$, so $25$ doesn't work.

If $n > 25$, the number must be divisible by $2^2 \cdot 3 \cdot 5 = 60$. But $\sqrt{60} > 7$, so the number would need a factor of $7$ also. Now we're looking at $2^2 \cdot 3 \cdot 5 \cdot 7 = 420$, which means we need $2^4 \cdot 3^2 \cdot 5 \cdot 7 \cdot 11 \cdot 13 \cdot 17 \cdot 19 = 232792560$, which means ... :)