Numerical bases and prime number theory

Question

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

Answer 1

If $N=3^x5^y7^z$ then $N$ has $(x+1)(y+1)(z+1)$ divisors.

$5N = 3^x5^{y+1}7^z$ has $(x+1)(y+2)(z+1)$ divisors

And $27N = 3^{x+4}5y7^z$ has $(x+5)(y+1)(z+1)$ divisors.

It may be easier to replace $x+1 =j; y+1=k; z+1=m$

So

So $(x+1)(y+2)(z+1)-(x+1)(y+1)(z+1)= 8$. So $j(k+1)m - jkm=jkm +jm - jkm = jm=8$.

And $(x+4)(y+1)(z+1)- (x+1)(y+1)(z+1)= = 18$.

So $(j+3)km - jkm = 3km =18$ or $km=6$

So $(x+1)(z+1) = jm= 8$ and $(y+1)(z+1)= km=6$.

So $z+1=m$ is a common divisor of $8,6$ so $m = 1,2$.

So we have either $z+1=1$ and $z=0$ and $y+1=6$ and $y=5$ and $x+1=8$ and $x=7$ and $N = 3^7*5^5$ which has $48$ divosors while $5N=3^7*5^6$ has $56$ divisors and $27N=3^{10}5^5$ has $66$ divisors.

(But presumably $z > 0$ otherwise they wouldn't have specified that $7$ was a factor. So I suppose the text isn't counting this as an answer although it fits all the requirements)

Or

$z+1=2$ and $z=1$ and $y+1=3$ and $y=2$ and $x+1=4$ and $x =3$ and $N=3^3*5^2*7=4725$ which has $24$ divisors while $5N = 3^3*5^3*7$ has $32$ divisors and $27N= 3^6*5^2*7$ has $42$ divisors.

Answer 2

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.

Answer 3

HINT.-$$N=3^x\cdot5^y\cdot7^z\quad\text{ has (x+1)(y+1)(z+1) divisors }\\5N=3^x\cdot5^{y+1}\cdot7^z\quad\text{ has (x+1)(y+2)(z+1) divisors }\\27N=3^{x+3}\cdot5^{y}\cdot7^z\quad\text{ has (x+4)(y+1)(z+1) divisors }$$ You can get from this the system $$(x+1)(z+1)[(y+2)-(y+1)]=8\\(y+1)(z+1)[(x+4)-(x+1)]=18$$ Then.........