Smallest integer N such that among every ten consecutive larger integers is at least one having at least three distinct prime factors? The problem is:

There exists an integer $N$ such that for any $n>N$, there exists $m \in \{n,n+1, \ldots ,n+9\}$ such that $m$ has at least $3$ distinct prime factors.

2 Years ago, My professor said to me it could be solved by using elementary number theory.
I considered this problem for 1 week. But I couldn't solve this problem. But professor didn't answer.
P.S I was considered about multiple of 6 2years ago. But it is not simple.
 A: I probably have most of it, maybe someone else will know how to finish it. I think you can do this by considering only the primes 2,3,5, in any detail.
You have, in particular, five consecutive even numbers. At least one of those is divisible by 3 as well. Unless this is the middle of five, there is another one divisible by 3. In which case you have, assuming at most two distinct prime factors, $ 2^a 3^b - 2^c 3^d = \pm 6.    $ So one of $a,c$ is one, and one of $b,d$ is one. The possibilities are $2^\alpha 3^\beta - 1 = \pm 1$ or $2^\alpha - 3^\beta = \pm 1.$ Quite finite, by Catalan.
Alright, so now the middle even number is also divisible by 3. One of the five even numbers is divisible by 5. Not the same one, that would make three prime factors. So we get
$$ 2^a 3^b - 2^c 5^d \in \{ \pm 2, \pm 4. \}     $$
Divide by the gcd which is 2 or 4, we get
$$  3^\beta - 2^\gamma 5^\delta = \pm 1,    $$
$$ 5^\delta - 2^\gamma 3^\beta  = \pm 1.    $$
The largest solution to any of the four equations summarized above, that I find, is $81 - 80 = 1,$ and I imagine this can be proved. Multiplying back by 2 or 4, the biggest original would be $324 - 320 = 4.$ 
So, if the middle even number is at least $324 + 6 = 330,$ so your $n$ is at least 325, the equations above do not have solutions, and so either the $2^a 3^b$ or the $2^c 5^d$ term has at least a third prime factor.  
