Is there an absolute constant $c>0$ such that, for all $1\le k<n$, the binomial coefficient $\left(\genfrac{}{}{0ex}{}{n}{k}\right)$ has a divisor in $(cn,n]$?

Erdős 387 (Schinzel) is OPEN; this is not a settlement. Rigorous contribution: a clean reduction. PROVEN (Legendre/Kummer): for 1≤k≤n/2 every prime in (n−k,n] divides C(n,k), so whenever that window contain

Erdős 387 (Schinzel) is OPEN; I did not solve it. Rigorous contribution: a clean reduction. PROVEN (Legendre/Kummer): for 1≤k≤n/2 every prime in (n−k,n] divides C(n,k), so whenever that window contains a prime, C(n,k) has a divisor >n/2 — giving c=1/2 unconditionally for all but the small-k-in-a-prime-free-window regime, which is the residual open core. Strong exact-computational evidence that the conjecture is TRUE with c=3/4 (global worst ratio D(n,k)/n is exactly 3/4, attained uniquely at n=4,k=2; per-n worst →1; no decay for fixed k). The famous n=99215,k=15 example has a divisor at 0.978n, so it does not threaten 387.