3. Find the smallest real number $x$ such that $45 \%$ of $x$ and $24 \%$ of $x$ are natural numbers.
Answer: $\frac{100}{3}$.

First solution. Since $\frac{45}{100} x$ is a natural number, $x$ is a positive rational number. Let $x=\frac{m}{n}$, where the fraction $\frac{m}{n}$ is irreducible. The numbers $\frac{45}{100} \cdot \frac{m}{n}=\frac{9 m}{20 n}$ and $\frac{24}{100} \cdot \frac{m}{n}=\frac{6 m}{25 n}$ are integers, so $9 m$ is divisible by $n$, and since $m$ and $n$ are coprime, 9 is divisible by $n$. Therefore, $n$ can be 1, 3, or 9. Since 9 and 20 are coprime, and $9 m$ is divisible by 20, $m$ is divisible by 20.

Since $6 m$ is divisible by $n$, and $m$ and $n$ are coprime, 6 is divisible by $n$. Since $6 m$ is divisible by 25, and 6 and 25 are coprime, $m$ is divisible by 25. Thus, $n \leqslant 3$ (only the cases $n=3$ and $n=1$ remain), and $m$ is divisible by 100, so $m \geqslant 100$. Therefore, $x \geqslant \frac{100}{3}$, and $x=\frac{100}{3}$ works, since $\frac{45}{100} x=15$ and $\frac{24}{100} x=8$.

Second solution. Note that $x$ must be positive, otherwise any percentage of it would be non-positive and could not equal a natural number. Let the natural numbers $a$ and $b$ be $45\%$ of $x$ and $24\%$ of $x$, respectively. Then the integer $c=2 b-a$ will be $2 \cdot 24-45=3$ percent of $x$. Since $3\%$ of $x$ is a positive number, $c \geqslant 1$, but then $x=\frac{100}{3} c \geqslant \frac{100}{3}$. The number $x=\frac{100}{3}$ works, since $\frac{45}{100} x=15$ and $\frac{24}{100} x=8$.

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