Is it true that for each $n\geq 2$ there are two primes $p, q$ such that (at least) one of them divides $\binom{n}{k}$ for each $1\leq k\leq n-1$?
Examples:
For $n=6: \binom{6}{1}=6; \binom{6}{2}=15; \binom{6}{3}=20; \binom{6}{4}=15; \binom{6}{5}=6.$ So we can have $p=2$ and $q=3$.
For $n=3: \binom{3}{1}=3; \binom{3}{2}=3.$ So we can have $p=3$ and $q=17$ (or any other prime). We just have to show that each $\binom{n}{k}$ is divisible by at least one of two primes.