There are no 120 consecutive numbers which all have exactly 120 divisors

Prove that the term $n=120$ in is $0$:

$24k$ with $k$ coprime to $6$ has exactly $120$ divisors

than

$k$ has exactly $15$ divisors

than

$k$ is a square number

than

$k$ cannot be $== 5, 7, 11 mod 12$ (since $5, 7, 11$ are not quadratic residues $mod 12$)

And we have a theorem:

A number $== 120, 168, 264 mod 288$ cannot have exactly $120$ divisors (since such numbers can be written as $24*k$ with $k$ coprime to $6$ and $k == 5, 7, 11 mod 12$)

So we have:

There are $120$ consecutive integers with exactly $120$ divisors

than

the start number must be $== 0, 265, 266, 267, 268, 269, 270, 271, 272, 273, 274, 275, 276, 277, 278, 279, 280, 281, 282, 283, 284, 285, 286, 287 mod 288$

than

the start number must be $== 0, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31 mod 32$

than

there are $4$ consecutive multiples of $32$ among these $120$ integers

than

one of these $4$ numbers must be == $64 mod 128$

than

the number of divisors of this number must be divisible by $7$ and cannot be $120$

And we have a contradiction, thus the proof is done.