11.4. Find the smallest natural number that has exactly 55 natural divisors, including one and the number itself.
Answer: $2^{10} \cdot 3^{4}$. Solution. If a natural number $n$ has the form $n=p_{1}^{k_{1}} \cdot p_{2}^{k_{2}} \cdot \ldots \cdot p_{m}^{k_{m}}$, where $p_{1}, p_{2}, \ldots, p_{m}$ are its distinct prime divisors, then the number of natural divisors of $n$ is $\left(k_{1}+1\right)\left(k_{2}+1\right) \ldots\left(k_{m}+1\right)$ (this follows from the general form of the divisor $p_{1}^{i_{1}} p_{2}^{i_{2}} \ldots p_{m}^{i_{m}}$ of the number $n, 0 \leq i_{1} \leq k_{1}$, $\left.0 \leq i_{2} \leq k_{2}, \ldots, 0 \leq i_{m} \leq k_{m}\right)$. Then $\left(k_{1}+1\right)\left(k_{2}+1\right) \ldots\left(k_{m}+1\right)=55=5 \cdot 11$, and therefore either $m=1$, or $m=2$. In the first case $k_{1}=54$ and $n=p_{1}^{54}$. In the second case $k_{1}=4, k_{2}=10$, i.e., $n=p_{1}^{4} \cdot p_{2}^{10}$. The smallest prime numbers are 2 and 3. Therefore, it remains to compare three numbers; $2^{54}, 2^{4} \cdot 3^{10}$ and $3^{4} \cdot 2^{10}$. Obviously, the smallest of them is $2^{10} \cdot 3^{4}$.