1. Determine the smallest natural number whose half is a perfect square, a third is a perfect cube, and a fifth is a perfect fifth power.
Solution. The Baraniot number must be divisible by 2, 3, and 5 (from the condition), so it must be of the form $n=2^{x} 3^{y} 5^{z}$. If it has other factors, it will not be the smallest. From the condition of the problem, we have:
$\frac{n}{2}=2^{x-1} 3^{y} 5^{z}$
(1) $\frac{n}{3}=2^{x} 3^{y-1} 5^{z}$
$\frac{n}{5}=2^{x} 3^{y} 5^{z-1}$

From (1) it follows that $x-1, y, z$ must be even. From (2) it follows that $x, y-1, z$ must be divisible by three. From (3) it follows that $x, y, z-1$ must be divisible by five.

Thus, $x$ is odd, divisible by 3 and by 5, $y$ is even, divisible by 5, and gives a remainder of 1 when divided by 3, $z$ is even, divisible by 3, and gives a remainder of 1 when divided by 5. Since $\mathbf{n}$ is the smallest, then $x, y$, and $z$ must also be the smallest, so we have $x=15, y=15, z=6$.

Thus, $n=2^{15} 3^{10} 5^{6}=30233088000000$.