Aakhnov $H . X$.

On the board, the expression $\frac{a}{b} \cdot \frac{c}{d} \cdot \frac{e}{f}$ is written, where $a, b, c, d, e, f$ are natural numbers. If the number $a$ is increased by 1, then the value of this expression increases by 3. If in the original expression the number $c$ is increased by 1, then its value increases by 4; if in the original expression the number $e$ is increased by 1, then its value increases by 5. What is the smallest value that the product $b d f$ can have?
The product of the denominators can be equal to 60. One of the possible examples: $\frac{20}{3} \cdot \frac{15}{4} \cdot \frac{12}{5}$.

Evaluation. Let the value of the original expression be $A$. Then, after the first operation, the product will take the value $\frac{a+1}{a} \cdot A=A+3$, from which

$A=3a$. This means that $A$ is a natural number divisible by 3. Similarly, it can be proven that $A=4c=5$. . Further reasoning can proceed in different ways.

First method. It is clear that $A$ is divisible by $3 \cdot 4 \cdot 5=60$, so $A \geq 60$.

Rewriting the equality $\frac{a}{b} \cdot \frac{c}{d} \cdot \frac{e}{f}=A$ as $\frac{A}{3b} \cdot \frac{A}{4d} \cdot \frac{A}{5f}=A$, we get $60bdf=A^2 \geq 60^2$, from which $bdf \geq 60$.

Second method. We have $\frac{1}{b} \cdot \frac{c}{d} \cdot \frac{e}{f}=3, \quad \frac{a}{b} \cdot \frac{1}{d} \cdot \frac{e}{f}=4, \quad \frac{a}{b} \cdot \frac{c}{d} \cdot \frac{1}{f}=5$. Multiplying the first equation by the second and dividing by the third, we get $\frac{e^2}{bdf}=\frac{12}{5}$. Since the fraction on the right is irreducible, the denominator $bdf$ is divisible by 5. Similarly, it can be proven that it is divisible by 3 and by 4, from which it follows that it is divisible by 60, that is, not less than 60.

## Answer

60.