Past exam of the mathematics course of the University of Cambridge / 2021 / iii / Paper 150 / 1 / b / Solution

Putby part a. Double-counting divisibility gives the first moment of the prime omega function:Moreover,soExpanding the square and using the given bound $\pi (x)\ll x/\log x$ now gives the Turán normal-order theorem for distinct prime divisors estimate

By the Chebyshev inequality, the number of $n\le x$ for whichis $O(x/L^{1/2})=o(x)$. Discard the $O(\sqrt{x})$ integers below $\sqrt{x}$. For $\sqrt{x}<n\le x$,and $(\log \log n{)}^{3/4}\sim L^{3/4}$. Hence, for all sufficiently large $x$, every remaining integer counted in the question also satisfies the preceding inequality. Therefore