The class received the following task. Determine the smallest natural number $H$ such that its half is equal to a perfect cube, its third is equal to a perfect fifth power, and its fifth is equal to a perfect square.

Pista and Sanyi were absent that day, and one of their friends told them the problem from memory, so Pista determined the smallest natural number $P$ with the (1) property, and Sanyi determined the smallest natural number $S$ with the (2) property, where $A, B, C, D, E, F$ are natural numbers.

$$
\begin{array}{lll}
P / 2=A^{2}, & P / 3=B^{3}, & P / 5=C^{5} \\
S / 2=D^{5}, & S / 3=E^{2}, & S / 5=F^{3}
\end{array}
$$

Which is the smallest among the numbers $H, P, S$?
I. In the homework problem, the number $H$ to be determined is such that its half, third, and fifth are natural numbers, so $H$ is divisible by 2, 3, and 5:

$$
H=2^{\alpha} \cdot 3^{\beta} \cdot 5^{\gamma} \cdot H_{1}
$$

where $\alpha, \beta, \gamma, H_{1}$ are natural numbers, and $H_{1}$ is not divisible by 2, 3, or 5. According to (3), the half, third, and fifth of $H$ are respectively

$$
\frac{H}{2}=2^{\alpha-1} \cdot 3^{\beta} \cdot 5^{\gamma} \cdot H_{1}, \quad \frac{H}{3}=2^{\alpha} \cdot 3^{\beta-1} \cdot 5^{\gamma} \cdot H_{1}, \quad \frac{H}{5}=2^{\alpha} \cdot 3^{\beta} \cdot 5^{\gamma-1} \cdot H_{1}
$$

thus $H / 2$ can only be a cube if $\alpha-1, \beta, \gamma$ are divisible by 3, $H / 3$ can only be a fifth power of a natural number if $\alpha, \beta-1, \gamma$ are divisible by 5, and finally $H / 5$ can only be a square if $\alpha, \beta$ are even and $\gamma$ is odd.

$H$ is clearly the smallest when $\alpha, \beta, \gamma, H_{1}$ are as small as possible. Thus $H_{1}=1$, and it is sufficient to determine the smallest $\alpha, \beta, \gamma$ that satisfy the above conditions. $\alpha$ must be divisible by 2 and 5, so it must also be divisible by 10; $\beta$ must be divisible by 6, and $\gamma$ must be divisible by 15. The smallest such values are

$$
\alpha=10, \quad \beta=6, \quad \gamma=15
$$

which also satisfy the additional requirements ( $\alpha-1$ is divisible by 3, $\beta-1$ is divisible by 5, and $\gamma-1$ is even). Therefore, the smallest value of $H$ is:

$$
H=2^{10} \cdot 3^{6} \cdot 5^{15}
$$

II. Similarly, the smallest solutions for (1) and (2) are:

$$
P=2^{15} \cdot 3^{10} \cdot 5^{6}, \quad S=2^{6} \cdot 3^{15} \cdot 5^{10}
$$

From (4) and (5),

$$
\begin{aligned}
\frac{H}{S} & =\frac{2^{4} \cdot 5^{5}}{3^{9}}=\frac{5}{3} \cdot\left(\frac{10}{9}\right)^{4}>1, & \text { so } & S<H \\
\frac{P}{S} & =\frac{2^{9}}{3^{5} \cdot 5^{4}}=\frac{2}{3} \cdot\left(\frac{4}{15}\right)^{4}<1, & \text { so } & P<S
\end{aligned}
$$

and thus the smallest of the numbers $H, P, S$ is $P$.