72nd Czech and Slovak Mathematical Olympiad

Given the decomposition $720=2^4\cdot 3^2\cdot 5$, the number $720$ has exactly three prime factors: $2$, $3$ and $5$. So, each of its factors is of the form $2^{\alpha }\cdot 3^{\beta }\cdot 5^{\gamma }$, where $\alpha ,\beta ,\gamma$ are non-negative integers (satisfying the inequalities $\alpha \le 4$, $\beta \le 2$ and $\gamma \le 1$ which we will not need further). Surely also the product of any three factors of $720$ is of the form $2^{\alpha }\cdot 3^{\beta }\cdot 5^{\gamma }$ with non-negative integers $\alpha ,\beta$ and $\gamma$. Considering any two numbers of this form, the first is a quotient of the second if and only if the values $\alpha ,\beta ,\gamma$ of the first number do not exceed the corresponding values of the second number.

We prove the problem statement by contradiction. Let us admit that some four factors of the number $720$ have the property that none of them divides the product of the other three factors. Then each of them contains in its prime decomposition some of the primes $2$, $3$, $5$ to a higher power than it has in its decomposition the product of the other three factors, and therefore any one of them. But there are four divisors and only three prime numbers, and this is the contradiction.

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Alternative solution.

We present one of several possible variations of a direct proof. We use the observations contained in the first paragraph of the previous solution.

Let us choose any four factors of the number $720$, call them Numbers. First, we choose three Numbers, that contain prime factor $2$ in powers not exceeding that power of $2$ in the fourth Number (if there are more than one such choice, we choose one of them). Then we select two of these three Numbers that contain prime factor $3$ in powers, that do not exceed power of $3$ of the third Number. Of these two Numbers, we finally select the one that contains the prime factor $5$ in a power not exceeding power of $5$ of the second Number. In the last selected number, each $p\in \{2,3,5\}$ has a power that does not exceed at least one of the powers of $p$ in the other three Number. This guarantees that the last selected Number has the property required by the problem statement.