7.6. Let $a, b, c$ be natural numbers, and the product $ab$ is divisible by $5c$, the product $c$ is divisible by $13a$, and the product $ca$ is divisible by $31b$. Find the smallest possible value of the product $abc$. Justify your answer.
Solution: From the condition that $ab$ is divisible by $5c$, it follows that at least one of the numbers $a$ and $b$ is divisible by 5. If this number is $a$, then from the condition that $bc$ is divisible by $13a$, it follows that one of the numbers $b$ or $c$ is also divisible by 5; if this number is $b$, then one of the numbers $a$ or $c$ is also divisible by 5 (due to the divisibility of $ca$ by $31b$). Thus, at least two of the numbers $a, b, c$ are divisible by 5. Similarly, at least two of these numbers are divisible by 13 and two by 31. Therefore, the product $abc$ is divisible by $5^2 \cdot 13^2 \cdot 31^2 = 2015^2 = 4060225$. Hence, $abc \geq 4060225$. The example $a = 5 \cdot 31 = 155, b = 5 \cdot 13 = 65, c = 13 \cdot 31 = 403$ shows that the obtained estimate is achievable.

Answer: $5^2 \cdot 13^2 \cdot 31^2 = 2015^2 = 4060225$.

Recommendations for checking:

| is in the work | points |
| :--- | :--- |
| Correct and fully justified answer | 7 points |
| Noted that the answer should be divisible by 9 and proved that the number in the answer is no more than 45, while the correct answer may be missing | 3 points |
| Provided the correct answer, supported by examples (in any quantity), but the uniqueness of the answer is not proven | 1 point |
| Answer without justification and/or incorrect answer | 0 points |