There are no 120 consecutive numbers which all have exactly 120 divisors
This article has been proposed for deletion. Please assess the validity of this proposal. To discuss this page in more detail, feel free to use the talk page. |
It has been suggested that this page be renamed. In particular: house style To discuss this page in more detail, feel free to use the talk page. |
This article needs to be tidied. In particular: Please take note of the message I posted concerning $\LaTeX$. It gives the maintenance admin team less tedious work to do. Please fix formatting and $\LaTeX$ errors and inconsistencies. It may also need to be brought up to our standard house style. To discuss this page in more detail, feel free to use the talk page. When this work has been completed, you may remove this instance of {{Tidy}} from the code. |
This article needs to be linked to other articles. You can help $\mathsf{Pr} \infty \mathsf{fWiki}$ by adding these links. To discuss this page in more detail, feel free to use the talk page. When this work has been completed, you may remove this instance of {{MissingLinks}} from the code. |
Prove that the term $n=120$ in This sequence is A072507 in the On-Line Encyclopedia of Integer Sequences (N. J. A. Sloane (Ed.), 2008).
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.
$\blacksquare$