## Task 4 - 251224

Determine all positive integers $n$ that have the following property: In the closed interval $\left[2^{n}, 2^{n+1}\right]$, there is at least one natural number divisible by $n^{3}$.
Due to $2^{3}=8<16=2^{4}, 2^{4}=16<27=3^{3}, 2^{5}=32<64=4^{3}, 2^{6}=64<125=5^{3}, 2^{7}=128<216=$ $6^{3}$ and $2^{8}=256<343=7^{3}$, it follows that $n=1$ (which is obviously a solution) or $n \geq 8$.

We show in the following that for $n \geq 8$ the inequality $n^{3} \leq 2^{n+1}$ holds:

It is $8^{3}=2^{9}=2^{8+1}$, so the inequality is satisfied for $n=8$. And if the inequality holds for a natural number $n \geq 8$, then it also holds for its successor $n+1$ due to $2^{(n+1)+1}=2 \cdot 2^{n+1} \geq 2 \cdot n^{3}=n^{3}+n^{3} \geq n^{3}+8 n^{2}=n^{3}+3 n^{2}+5 n^{2} \geq$ $n^{3}+3 n^{2}+40 n \geq n^{3}+3 n^{2}+3 n+1=(n+1)^{3}$, and thus for all natural numbers $n \geq 8$.

It follows that for all $n \geq 8$, the value $n^{3}$ lies directly in the interval $\left[2^{n} ; 2^{n+1}\right]$, and thus satisfies the condition of the problem, or $n^{3}<2^{n}$ holds. In this case, however, among any $n^{3}$ consecutive natural numbers, exactly one is divisible by $n^{3}$, and in particular one of the numbers $2^{n}+1,2^{n}+2, \ldots$, $2^{n}+n^{3}<2^{n}+2^{n}=2^{n+1}$, which all lie in the interval $\left[2^{n} ; 2^{n+1}\right]$.

Thus, it follows that exactly those positive integers $n$ that satisfy $n=1$ or $n \geq 8$ are solutions to the problem.