Find the largest number that $ n(n^2-1)(5n+2) $ is always divisible by? My Solution:
$$ n(n^2-1)(5n+2) = (n-1)n(n+1)(5n+2) $$

*

*This number is divisible by 6 (as at least one of 2 consecutive integers is divisible by 2 and one of 3 consecutive integers is divisible by 3.


*$ 5n+2 \equiv 5n \equiv n \mod 2 $ then $n$ and $5n+2$ have the same pairness and at least one of $n+1$ and $5n+2$ is divisible by 2.


*$ n \equiv 5n \equiv 5n+4 \mod 4 \to $
if $ 2\ | \ n+1 \to n - 1 $ or $ n + 1 $ is divisible by 4
if $ 2\ | \ 5n+2 \to n $ or $ 5n + 2 $ is divisible by 4
The expression is divisible by 6 and has 2 even integers and one of them is divisible by 4 $\to$ is divible by 24.

Alternatively.
Let $p$ be an odd prime.  If $p|n$ but $p^2 \not \mid n$ then we have $p\not \mid n-1, n-2, 5n + 2$ so we need never have any odd prime $p^2|n(n^2-1)(5n+2)$.
If $p=3$ we must have one of $n-1, n, n+1$ be divisible by $3$ so we must have $3|n(n^2-1)(5n+2)$ but we need not have $3^2|n(n^2-1)(5n+2)$.
If $p=5$ we will not have $5|5n+2$ and as we have $5$ options for $n \pmod 5$ we can avoid $n \equiv 0, 1, 4$ (by having $n\equiv 2,3\pmod 5$ and thus have $n, n+1, n-1\not \equiv 0\pmod 5$ and so $5\not \mid n(n^2-1)(5n+2)$.
If $p > 5$ we can have more than $4$ options for $n\pmod 5$ and so we can have $n \not \equiv 0, 1,-1, -2\cdot 5^{-1} \pmod p$. ANd that way we can avoid $p|n(n^2-1)(5n+2)$.
So no odd prime other than $3$ need divide $n(n^2-1)(5n+2)$ and $3^2$ need not divide it.
And as what power of $2$ must divide $n(n^2-1)(5n+2)$.
One of $n,n+1$ must be even so $2$ must divide.  If $n$ is odd then $n-1$ and $n+1$ is even and so $4$ must divide.  But one of $n-1$ or $n+1$ must be divisible by $4$ so $8$ must divide.
If $n$ is even then $5n+2$ is even so $4$ must divide.  If $n$ is not divisible by $4$ then $n = 4k + 2$ for some $k$ and $5n+2 = 20k + 12$ is divisible by $4$ and so $8$ must divide.  And if $n$ is divisible by $4$ then as $5n + 2$ is even $8$ must divide.
That exhausts all cases so $8$ must divide $n(n^2-1)(5n+2)$.
Now no higher power of $2$ must divide as we can chose $n$ odd so that $4$ but not $8$ divides one of $n\pm 1$.  For example if $n = 5$ and so $n+1 = 6$ and $n-1 =4$ and only $8$ need divide.  Or we can choose an even $k$ where $n = 4k + 2$ and $5n+2 = 20k + 12= 4(5k + 3)$ is divisible by $4$ but not by $8$.
So we must have $3$ and $8$ divide $n(n^2-1)(5n+2)$ but no other prime or higher power of $3$ or higher power of $2$ need divide.
SO $24$ is the largest number that must divide.
