# 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:
$$0 = frac{0}{n} < frac{| X_{M,k} cap {0,1,cdots,n} | }{n} < frac{n}{n} = 1$$