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 $m\geq 5$.
Best Answer
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.