8
$\begingroup$

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..

$\endgroup$
4
$\begingroup$

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., $420$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.