Apply part a to $A=\{1,\ldots,\lfloor x\rfloor\}$ and all primes. Since
$$
|\{n\leq x:d\mid n\}|=\left\lfloor\frac xd\right\rfloor,
$$
we obtain
$$
S(A,\mathbb P,z)
=x\sum_{d\mid P(z)}\frac{\mu(d)}d+O\left(\sum_{d\mid P(z)}1\right)
=x\prod_{p\leq z}\left(1-\frac1p\right)+O(2^{\pi(z)}).
$$
For $z\leq\log x$, the error is at most $2^z\leq x^{\log2}=o(x/\log z)$. By <Mertens theorem>, as $z\to\infty$,
$$
\prod_{p\leq z}\left(1-\frac1p\right)
=\frac{e^{-\gamma}+o(1)}{\log z}.
$$
Thus
$$
|\{n\in[1,x]:n\text{ has no prime factor at most }z\}|
=\left(e^{-\gamma}+o(1)\right)\frac{x}{\log z},
$$
so one may take $C=e^{-\gamma}>0$. For bounded $z$, the preceding exact product formula gives the corresponding fixed density.

Solved by gpt-5.6-sol high.

Solution

Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 150 / 1 / b / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2025/iii/paper-150/1/b/solution