Numerical bases and prime number theory Let $N = 3^x$. $5^y$. $7^z$. Find $N$ such that $5N$ and $27N$ have $8$ and $18$ dividers, respectively, more than $N$.
I did what the statement asks and it was here:
dividers: $(x + 1) (y + 2) (z + 1) = (x + 1) (y + 1) (z + 1) + 8; (x + 4) (y + 1) (z + 1) = (x + 1) (y + 1) (z + 1) + 18$
System

As Steve Kass's question comment suggests, you have
$$\begin{equation}\begin{aligned}
(x + 1) (y + 2) (z + 1) & = (x + 1) (y + 1) (z + 1) + 8 \\
((y + 1) + 1)\left((x + 1) (z + 1)\right) & = (x + 1) (y + 1) (z + 1) + 8 \\
(x + 1) (y + 1) (z + 1) + (x + 1)(z + 1) & = (x + 1) (y + 1) (z + 1) + 8 \\
(x + 1)(z + 1) & = 8
\end{aligned}\end{equation}\tag{1}\label{eq1A}$$
$$\begin{equation}\begin{aligned}
(x + 4) (y + 1) (z + 1) & = (x + 1) (y + 1) (z + 1) + 18 \\
((x + 1) + 3)\left((y + 1) (z + 1)\right) & = (x + 1) (y + 1) (z + 1) + 18 \\
(x + 1) (y + 1) (z + 1) + 3(y + 1)(z + 1) & = (x + 1) (y + 1) (z + 1) + 18 \\
3(y + 1)(z + 1) & = 18 \\
(y + 1)(z + 1) & = 6
\end{aligned}\end{equation}\tag{2}\label{eq2A}$$
Since the factors are positive integers and $z + 1$ is a factor in both \eqref{eq1A} and \eqref{eq2A}, then $z + 1$ must divide $\gcd(8,6) = 2$. Thus, $z + 1 = 1$ or $z + 1 = 2$. For the first case, this gives $z = 0$, $x + 1 = 8 \implies x = 7$ and $y + 1 = 6 \implies y = 5$. I'll leave the $z + 1 = 2$ case for you to do.
