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

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.
