\section*{Exercise 1 - 321021}

Determine the smallest natural number \(z\) that has exactly four divisors \(t_{1}, t_{2}, t_{3}, t_{4}\) with \(1<t_{1}<t_{2}<t_{3}<\) \(t_{4}<z\)!
}

If the natural number \(z\) has the prime factorization

\[
z=p_{1}^{a_{1}} \cdot p_{2}^{a_{2}} \cdot \ldots \cdot p_{n}^{a_{n}}
\]

where the \(p_{i}(i=1,2, \ldots, n)\) are distinct prime factors and the \(a_{i}\) are their frequencies in the factorization, then the number of divisors \(t\) of \(z\) (including 1 and \(z\) itself) is given by

\[
t=\left(a_{1}+1\right) \cdot\left(a_{2}+1\right) \cdot \ldots \cdot\left(a_{n}+1\right)
\]

Since the number \(z\) is supposed to have 6 divisors (including 1 and \(z\)), we have

\[
6=\left(a_{1}+1\right) \cdot\left(a_{2}+1\right) \cdot \ldots \cdot\left(a_{n}+1\right)
\]

Each factor \(\left(a_{i}+1\right)\) is greater than 1, so the only possible product is \(6=3 \cdot 2\), meaning \(z\) has exactly 2 prime factors, one of which appears twice and the other once, i.e.,

\[
z=p_{1}^{2} \cdot p_{2}
\]

If we substitute the smallest prime numbers 2 and 3 for \(p_{1}\) and \(p_{2}\), we get \(z=2^{2} \cdot 3=12\). 12 is the number we are looking for. Its divisors, excluding 1 and \(z\) itself, are: 2, 3, 4, and 6.