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

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.........
