Team Selection Test

The answer is $420$.

Let us consider the positive integer $m$ so that $m^3\le n<(m+1{)}^3$. As $n=420$ satisfies the conditions, we will consider the case when $m\ge 7$. Note that each of $m$, $m-1$, $m-2$ and $m-3$ divides $n$ and hence $\mathrm{lcm}(m,m-1,m-2,m-3)$ divides $n$. Since $\gcd (n-1,n-2)=1$, $\gcd (n,n-3)$ divides $3$ and $\gcd (n(n-3),(n-1)(n-2))$ divides $2$, we have that

$$\frac{m(m-1)(m-2)(m-3)}{6}$$ divides $n$. Therefore, $$\frac{m(m-1)(m-2)(m-3)}{6}<(m+1{)}^3$$ and hence $m\le 12$.

If $m=11$ or $12$, then $11\cdot 7\cdot 5\cdot 9\cdot 8=27720\mid n$, but $13^3=2197<27720$.

If $m=9$ or $10$, then $7\cdot 5\cdot 9\cdot 8=2520\mid n$, but $11^3=1331<2520$.

If $m=8$, then $7\cdot 5\cdot 3\cdot 8=840\mid n$, but $9^3=729<840$.

If $m=7$, then $7\cdot 5\cdot 3\cdot 4=420\mid n$ and $n<8^3=512$. Therefore $n=420$.