}

It is \(82944000000=2^{16} \cdot 3^{4} \cdot 5^{6}\), so that only the prime numbers 2, 3, and 5 appear in the prime factorization of the \(x_{i}\). We can thus write \(x_{i}=2^{a_{i}} \cdot 3^{b_{i}} \cdot 5^{c_{i}}\) with non-negative integers \(a_{i}, b_{i}, c_{i}\) for \(i \in\{1,2,3,4\}\).

Then, according to a) \(a_{1}+\cdots+a_{i}=16, b_{1}+\cdots+b_{4}=4\) and \(c_{1}+\cdots+c_{4}=6\).

It is \(24=2^{3} \cdot 3^{1}\), so that according to b) all \(a_{i} \geq 3\) and all \(b_{i} \geq 1\). From the latter, it follows immediately with the sum condition for the \(b_{i}\) that \(b_{1}=\cdots=b_{4}=1\). Furthermore, at least one of the \(c_{i}\) must be equal to 0, as otherwise the gcd would also be divisible by 5. Similarly, at least one of the \(a_{i}\) must be exactly equal to 3.

It is \(120000=2^{6} \cdot 3^{1} \cdot 5^{4}\), so that according to c) all \(a_{i} \leq 6\) and all \(c_{i} \leq 4\), where these maximum values must also be assumed at least once. From the sum condition for the \(a_{i}\), these upper and the previously shown lower bound for their size, it follows that exactly one of the \(a_{i}\) must take the value 6, a second one the value 4, and the two remaining ones the value 3.

It is \(1200=2^{4} \cdot 3^{1} \cdot 5^{2}\), so that according to d) \(a_{1} \geq 4, a_{2} \geq 4, c_{1} \geq 2\) and \(c_{2} \geq 2\) follows. In particular, it follows that \(a_{3}=a_{4}=3\). Since at least one of the \(c_{i}\) has the value 4, the sum condition for the \(c_{i}\) can only be satisfied if \(c_{3}=c_{4}=0\), one of the two values \(c_{1}\) and \(c_{2}\) is equal to 4 and the other is equal to 2.

It is \(30000=2^{4} \cdot 3^{1} \cdot 5^{4}\), so that according to e) \(a_{2} \leq 4, a_{3} \leq 4, c_{2} \leq 4\) and \(c_{3} \leq 4\) follows, where at least one of the \(a\) or \(c\) values must also assume the bound. Due to the previously obtained estimate for \(a_{2}\), it follows that \(a_{2}=4\) and analogously because \(c_{3}=0\) also \(c_{2}=4\). It follows that \(c_{1}=2\) and finally \(a_{1}=6\).

\section*{Together we obtain}

\(x_{1}=2^{a_{1}} \cdot 3^{b_{1}} \cdot 5^{c_{1}}=2^{6} \cdot 3^{1} \cdot 5^{2}, \quad x_{2}=2^{a_{2}} \cdot 3^{b_{2}} \cdot 5^{c_{2}}=2^{4} \cdot 3^{1} \cdot 5^{4}\)

\(x_{3}=2^{a_{3}} \cdot 3^{b_{3}} \cdot 5^{c_{3}}=2^{3} \cdot 3^{1} \cdot 5^{0}, \quad x_{4}=2^{a_{4}} \cdot 3^{b_{4}} \cdot 5^{c_{4}}=2^{3} \cdot 3^{1} \cdot 5^{0}\).
\section*{Problem 2 - 111042}

All ordered quadruples \(\left(x_{1}, x_{2}, x_{3}, x_{4}\right)\) of positive integers are to be determined that have the following properties:

a) The product of these four numbers is equal to 82944000000.

b) Their greatest common divisor (gcd) is equal to 24.

c) Their least common multiple (lcm) is equal to 120000.

d) The greatest common divisor of \(x_{1}\) and \(x_{2}\) is equal to 1200.

e) The least common multiple of \(x_{2}\) and \(x_{3}\) is equal to 30000.