154. a) Find the number if the product of all its divisors is 108.

b) Find the number if the product of its divisors is 5832.
154. b) $5832=2^{3} \cdot 3^{6}$; therefore, the required number $A=2^{x} 3^{y}$, where $x<3 ; y<6$. Based on № 151, we find:

$$
\left(2^{x} \cdot 3^{y}\right)^{\frac{1}{2}(x+1)(y+1)}=2^{3} \cdot 3^{6}
$$

From the system $\left\{\begin{array}{l}\frac{x(x+1)(y+1)}{2}=3 \\ \frac{y(x+1)(y+1)}{2}=6\end{array}\right.$, by dividing we find $y=$ $=2 x$; substituting $y=2 x$ into the last equation of the system, we get $x(x+1)(2 x+1)=6$, hence $x=1, y=2$.

Answer. 18.