Prove there exist 135 consecutive integers that each have a perfect cube $>1$ that divides it.

Question

As of 22:13, the answer below does not satisfy the constraints of my question, so I'm still accepting answers.

---

Prove there exist 135 consecutive integers, $x_1, x_2, ... , x_{135}$, such that for each of these integers, $x_i$, there exists a number $k_i$ such that $k_i>1$ is a perfect cube where $k_i | x_i$.

---

I was told that I could use GCRT to help with this. I got stuck after rewriting the given statement as a system of linear congruences.

Answer 1

Chose $135$ distinct primes $p_0,\dots,p_{134}$.

Now consider the system $$X \equiv -i \pmod{p_i^3}$$ for $i=0, \dots, 134$.

Since the moduli are pairwise co-prime there exists a solution $x_0$.

It follows that $x_0 + i $ is divisible by $p_i^3$ for each $i$.