if $xy=2808, z(y+4)=2808$ find $y$ Let $x,y,z$ are postive integer,and such
$$\begin{cases}
x\ge 30\\
xy=2808\\
z(y+4)=2808
\end{cases}$$
Find $y$
I have take $2808=2\times 1404=2^3\times 3^3\times 13$

Since $y = \frac{2808}{x}$ and $x \geq 30$, we have that $y \leq \frac{2808}{30} = 93.6 \implies y \leq 93$.
Since $y$ divides $2808$, $y = 2^a \cdot 3^b$ or $y = 2^a \cdot 3^b \cdot 13$ for $0 \leq a, b \leq 3$.
Writing down such $y \leq 93$ sequentially is trivial but tedious, and gives the list:
$y \in \lbrace 1, 2, 3, 4, 6, 8, 9, 12, 13, 18, 24, 26, 27, 36, 52, 54, 72, 78 \rbrace$
Observe that $z(y+4) = 2808 \implies y+4$ divides $2808$, and that the only $y$ in the list such that $y+4$ dividies 2808 are $y = 2, 4, 8, 9$.
Then $x = \frac{2808}{y}$ and $z = \frac{2808}{y+4}$ gives solutions
$$(x, y, z) = (1404, 2, 468), \;(702, 4, 351), \; (351, 8, 234), \;(312, 9, 216)$$
