set theory – LIMIT PROBLEM FROM BRAZIL OLYMPIAD

Actually I found this problem at MSE and it was unanswered. I found it really challenging and I am really clueless how to continue. So here’s the original argument of the author.

Let $M,k$ be two positive integers. Define $X_{M,k}$ as the set of the numbers $p_1^{alpha_1}cdot p_2^{alpha_2} cdots p_r^{alpha_r}$ where $p_i$ are prime numbers such that $M leq p_1 < p_2 < cdots < p_r$ and $alpha_i geq k$. Prove that there are positive real numbers $beta(M,k)$ and $c(M,k)$ such that
$$
lim_{nrightarrow infty} frac{| X_{M,k} cap {0,1,cdots,n} | }{n^{beta(M,k)}} = c(M,k)
$$

and determine the value of $beta(M,k)$.



First let’s simplify what $X_{M,k}$ is…
$$
begin{align*}
X_{M,k} &= {p_1^{alpha_1}cdot p_2^{alpha_2} cdots p_r^{alpha_r} text{ | } M leq p_i < p_{i+1} text{ and } alpha_i geq k}\
&= {(p_1^k p_2^k cdots p_r^k) cdot p_1^{gamma_1}cdot p_2^{gamma_2} cdots p_r^{gamma_r} text{ | } M leq p_i < p_{i+1} text{ and } gamma_i geq 0}\
&= {s cdot p_1^{gamma_1}cdot p_2^{gamma_2} cdots p_r^{gamma_r} text{ | } M leq p_i < p_{i+1} text{ and } gamma_i geq 0} text{ where } s = p_1^k p_2^k cdots p_r^k\
end{align*}
$$

We can then see that $X_{M,k}$ is a proper subset of the set of multiples of $s$.
Therefore we can see that $|X_{M,k} cap {0,1,cdots,n}| < |{text{multiples of s} leq text{ then n}}| < n$.

Using that, and the fact that $beta(M,k) > 0$ we can conclude that:
$$
frac{| X_{M,k} cap {0,1,cdots,n} | }{n^{beta(M,k)}} < frac{n}{n^{beta(M,k)}} = n^{1-beta(M,k)}
$$

Now see that $| X_{M,k} cap {0,1,cdots,n} | > 0$ only when $n geq s$. Therefore, it’s also valid to say that for every $n geq s$ it follows that:
$$
0 = frac{ 0 }{n^{beta(M,k)}} < frac{| X_{M,k} cap {0,1,cdots,n} | }{n^{beta(M,k)}} < frac{n}{n^{beta(M,k)}} = n^{1-beta(M,k)}
$$

Finally, see that it’s possible to let $n^{1-beta(M,k)} = n^0 = 1$ if we take $beta(M,k) = |X_{M,k} cap {s} | = 1$.

Joining my conclusions till now, we can say that if we fix $beta(M,k) = |X_{M,k} cap {s} | = 1$ then for every $n geq s$ it’s valid that:
$$
0 = frac{0}{n} < frac{| X_{M,k} cap {0,1,cdots,n} | }{n} < frac{n}{n} = 1
$$