# Find the largest integer $n$ such that $n$ is divisible by all positive integers less than $\sqrt[3]{n}$

Find the largest integer $$n$$ such that $$n$$ is divisible by all positive integers less than $$\sqrt[3]{n}$$.

420 satisfies the condition since $$7<$$ $$\sqrt[3]{420}<8$$ and $$420=\operatorname{lcm}\{1,2,3,4,5,6,7\}$$

Suppose $$n>420$$ is an integer such that every positive integer less than $$\sqrt[3]{n}$$ divides $$n .$$

Then $$\sqrt[3]{n}>7,$$ so $$420=\operatorname{lcm}(1,2,3,4,5,6,7)$$ divides $$n$$ thus $$n \geq 840$$ and $$\sqrt[3]{n}>9 .$$

Thus $$2520=\operatorname{lcm}(1,2, \ldots, 9)$$ divides $$n$$ and $$\sqrt[3]{n}>13$$

now this pattern looks continues,but i am not able to prove that this pattern always continues..

You have the basic right idea of showing the $$\operatorname{lcm}$$ values increase faster than the cube of the highest value used in the $$\operatorname{lcm}$$, with the following being one way to finish the solution. Define for any positive integer $$m$$,

$$f(m) = \operatorname{lcm}(1,2,\ldots,m) \tag{1}\label{eq1A}$$

For some prime $$m \gt 8$$, consider if you have

$$f(m) \gt 8m^3 \tag{2}\label{eq2A}$$

For any integer $$n \ge m^3$$, since $$\sqrt[3]{n} \ge m$$, you would need to include $$m$$ in the $$\operatorname{lcm}$$. However, \eqref{eq2A} shows you actually need $$n \gt 8m^3$$, so $$\sqrt[3]{n} \gt 2m$$. The less restrictive formulation of Betrand's postulate states there's always at least one prime $$p$$ where $$m \lt p \lt 2m$$, so since $$p \gt 8$$ and this prime $$p$$ must be multiplied into the $$\operatorname{lcm}$$ value, you have

$$f(2m) \gt p(8m^3) \gt 8(8m^3) = 64m^3 \tag{3}\label{eq3A}$$

Thus, you actually have $$n \gt 64m^3$$, giving $$\sqrt[3]{n} \gt 4m$$. You can use the procedure above repeatedly, in an inductive type fashion, to get $$n \gt (8^{k})m^{3}$$ for $$k = 1, 2, 3, \ldots$$, showing there is no larger $$n$$ which works.

As for the base case, note that

$$f(11) = 27\text{,}720 \gt 10\text{,}648 = 8(11^3) \tag{4}\label{eq4A}$$

Since it seems you've checked the other cases for $$m \lt 11$$, this shows the largest $$n$$ which works is what you've already found, i.e., $$420$$.