5. Find the largest positive integer $x$ such that $x$ is divisible by all the positive integers $\leq \sqrt[3]{x}$.
5. The answer is $x=420$.

Let $p_{1}, p_{2}-p_{1}, \ldots$ be all the primes arranged in increasing order. By Bertrand' $\&$ Postulate, we have $p_{i}p_{k-1}^{3}$.
Let $p_{k} \leq \sqrt[1]{x}2^{3}$ and $11>3^{2}$, we have $2^{3} 3^{2} \mid x$. Since $k \geq 5, \operatorname{gcd}\left(p_{k} p_{k-1} p_{k} \cdot 2 \cdot 2^{3} 3^{2}\right)=1$ and thus $2^{3} 3^{2} p_{k} p_{k-1} p_{k-2} \mid x$. This means we have $x \geq 72 p_{k} p_{k-1} p_{k-2}>64 p_{k} p_{k-1} p_{k-2}>p_{k+1}^{3}$, implying $p_{k+1}7$. It then follows that $x$ is divisible by $2^{2} \cdot 3 \cdot 5 \cdot 7=420$.
Finally, suppose $\sqrt[3]{x} \geq 9$. We theu have $2^{3} \cdot 3^{2} \cdot 5 \cdot 7 \mid x$, i.c., $x \geq 2^{3} \cdot 3^{2} \cdot 5 \cdot 7=2520$, which is a contradiction since $x<1331$. Tluts $\sqrt[3]{x}<9$. or $x<729$. Since $420 \mid x$ aud $x<729$. we have $x=420$.
Alternatively. since $x<1331$ and $420 \mid x$, we only need to check the cases $x=$ 420.840 .1260 .