Given $0 < p < n$, prove there exists $n$ consecutive natural numbers such that each natural is divisible by at least $p$ distinct primes.

Question

Given $0 < p < n$, prove there exists $n$ consecutive natural numbers such that each natural is divisible by at least $p$ distinct primes.

Is there a general proof method to prove this statement?

I brushed up my knowledge on the Chinese Remainder Theorem and Euclid's Theorem since they seem relevant, but I cannot find the necessary insight.

Answer 1

We can find by the Chinese remainder theorem a $x$ such that:
$x\equiv -1\pmod{p_1\cdot p_2\cdots p_n}$
$x\equiv -2\pmod{p_{n+1}\cdot p_{n+2}\cdots p_{2n}}$
.
.
.
$x\equiv -n\pmod{p_{(n-1)\cdot n+1}\cdots p_{n\cdot n}}$
and the primes are chosen to be all different.
You can see easily that $x+1,x+2, \ldots x+n$ have each one of them at least $n$ prime factors which means at least $p$ prime factors.