Numbers that are divisible by the number of primes smaller than them

Solution 1:

With respect to your question if there is, for each $k$, an $n$ such that $\dfrac{n}{\pi(n)}=k$ I think the answer is affirmative. If we consider the function $f(x) = \dfrac{x}{\pi(x)}$, we can prove that $\displaystyle\lim_{x\to +\infty}f(x)=+\infty$ (by the prime number theorem or by more elementary bounds). If $f$ were continuous, as $f(2) = 2$, we should have an intermediate value such that $f(x)=k$. This naive approach doesn't work (as $f$ is not continuous). However, we can do something similar:

We know that there is an $n\in\mathbb{N}$ such that $f(n)<k$ and $f(n+1)\geq k$ (as $f$ tends to infinite, it can't always be bounded by $k$). If $f(n+1)=k$, then we're done. If not, $f(n+1)>k$. This two inequalities can be rewritten as $n < k \pi(n)$ and $n+1 > k\pi(n+1)$. This means that $$k \pi (n+1) < n+1 < k\pi(n) + 1$$ That is $k(\pi(n+1) - \pi(n)) < 1$. This implies that $\pi(n)=\pi(n+1)$. But then, $$k\pi(n+1) = k\pi(n) < n+1 < k\pi(n)+1$$ Absurd, since $n+1$ is an integer in between two consecutive integers.