Consistent number of divisors

The positive integer $l$ has $d(l)$ number of divisors. Let $N=2^x\times 3^y\times k$ where $\text{gcd}(2, k)=1=\text{gcd}(3,k)=1$ . We can neglet the number of divisors of $k$ , since it appears exactly once in each equation. From that $\begin{aligned} d(18N) & = d(24N) \\ d(2\times 3^2\times N) & = d(2^3\times 3\times N) \\ d(2^{x+1}\times 3^{y+2}) & = d(2^{x+3}\times 3^{y+1} \\ (x+2)(y+3) & = (x+4)(y+2) \\ xy+3x+2y+6 & = xy+2x+4y+8 \\ x=2y+2 \end{aligned}$ Since $d(18N)=d(27N)$ , using a similar way we get the following: $y+2=x$ By adding the two last equations: $\begin{aligned} x+y+2 & = x+2y+2 \\ 0 & = y \end{aligned}$ From that $x=2$ and $d(18N)=d(24N)=d(27N)=12$ Since $M>27$ , it is clear that the minimum value of $M$ is $2^{12-1-2}$ .

Therefore the answer is $\boxed{512}$ .