Cubes - mind gobblers

let the number be given by

$n=100*x+y$

with $1<=x<=9$ and $0<=y<=99$

then

$n^3=(10^2*x+y)^3$

$n^3=10^6*x^3+3*10^4*x^2y+3*10^2*x*y^2+y^3$

now to get the last two digits we simply take this $mod 100$ which simplifies down to

$n^3 = y^3$ $mod$ $100$ now last digit, $d$ , needs to satisfy $d^3 = 4$ $mod$ $10$ , and this is only true for $d=4$ . So now there are only 10 possible values for y, namely $(4,14,24,34,44,54,64,74,84,94)$ trying each of these we get that only

$14^3 = 44$ $mod$ $100$ and $64^3 = 44$ $mod$ $100$ .

Thus there are 2 possible values for y and 9 possible values of x giving $\boxed{18}$ total.