Solution

Past exam of the mathematics course of the University of Cambridge / 2025 / ii / Paper 4 / 11G / c / Solution

For every real $X\geq1$,
$$
P(X)\leq4^X.
$$
For integer $n$, strong induction proves this by splitting according to parity. If $n=2k$, every prime in $(k,2k]$ divides $\binom{2k}{k}$, so
$$
P(2k)\leq P(k)\binom{2k}{k}\leq4^k4^k.
$$
If $n=2k+1$, the <valuation of a near-central binomial coefficient> gives
$$
P(2k+1)\leq P(k+1)\binom{2k+1}{k+1}\leq4^{k+1}4^k.
$$
The binomial estimates follow from the binomial theorem; in the odd case the two equal central coefficients together are at most $2^{2k+1}$. The result for real $X$ follows by replacing $X$ with $\lfloor X\rfloor$.

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 Incoming links: Elementary primorial bound