does a number that contains all primes less than it exist?
This is impossible. For every $n>1$, we have gcd$(n,n-1)=1$. Since there is a prime dividing $n-1$, the result follows.
Edit : I should have assumed $n>2$ since I need a prime divisor of $n-1$.
To make Lord Shark's comment explicit:
Let $n$ be composite and let $p$ be its largest prime factor. Then certainly $n\ge 2p$. By Bertrand's postulate, there exists a prime $q$ between $p$ and $2p$, hence there exist primes below $n$ that do not divide $n$.