IrMO 1989 Problem 10

(i): Prove that if $n$ is a positive integer, then $$\binom{2n}{n}=\frac{(2n)!}{(n!)^2}$$ is a positive integer that is divisible by all prime numbers $p$ with $n<p\le 2n$ , and that $$\binom{2n}{n}<2^{2n}.$$ (ii): For $x$ a positive real number, let $\pi(x)$ denote the number of prime numbers $p \le x$ . [Thus, $\pi(10) = 4$ since there are $4$ primes, viz., $2$ , $3$ , $5$ , and $7$ , not exceeding $10$ .]Prove that if $n \ge 3$ is an integer, then (a) $$\pi(2n) < \pi(n) + {{2n}\over{\log_2(n)}};$$ (b) $$\pi(2^n) < {{2^{n+1}\log_2(n-1)}\over{n}};$$ (c) Deduce that, for all real numbers $x \ge 8$ , $$\pi(x) < {{4x \log_2(\log_2(x))}\over{\log_2(x)}}.$$