An integer's product of all its positive integer divisors is $2^{120} \cdot 3^{60} \cdot 5^{90}$. What could this number be?
Let the number we are looking for be denoted by $N$ ( $N$ can be negative). The prime factors of $N$ are those and only those prime numbers that appear in the product of all divisors of $N$. Therefore, $N=2^{l} \cdot 3^{m} \cdot 5^{n}(l, m, n$ are positive integers). Let $d(N)$ denote the number of all positive divisors of $N$, and $P$ the product of these divisors. If $N$ is not a perfect square, then every positive divisor of $N$ has a different positive divisor pair, the product of which is $|N|$, hence $P=|N|^{\frac{d(N)}{2}}$. If $N$ is a perfect square, then the pairing can also be performed, except for the positive divisor $\sqrt{|N|}$. In this case,

$$
P=|N|^{\frac{d(N)-1}{2}} \sqrt{|N|}=|N|^{\frac{d(N)}{2}}
$$

In every case, therefore,

$$
P=|N|^{\frac{d(N)}{2}}
$$

From the form $N=2^{l} \cdot 3^{m} \cdot 5^{n}$, it follows that $d(N)=(l+1)(m+1)(n+1)$, since the positive divisors of $N$ are the numbers $2^{i} \cdot 3^{j} \cdot 5^{k}$, where $0 \leqq i \leqq l, 0 \leqq j \leqq m, 0 \leqq k \leqq n$ are integers, and $i, j, k$ can independently take $l+1, m+1, n+1$ different values, respectively. Based on the above,

$$
2^{120} \cdot 3^{60} \cdot 5^{90}=\left(2^{l} \cdot 3^{m} \cdot 5^{n}\right)^{\frac{(l+1)(m+1)(n+1)}{2}}
$$

From this, it can be read that $l: m: n=120: 60: 90$, that is:

$$
\begin{aligned}
l & =4 t \\
m & =2 t(t \text { is a positive integer). } \\
n & =3 t
\end{aligned}
$$

(The exponents of the powers with the same base must match on both sides, since the prime factorizations of the same number cannot be different.)

If we now compare, for example, the exponents of the powers of 3 on both sides of equation (1), we get the equation

$$
t(4 t+1)(2 t+1)(3 t+1)=60
$$

which has only one solution in the set of positive integers:

$$
t=1
$$

(The expression on the left side strictly monotonically increases as $t$ increases.) According to this, $N=2^{4} \cdot 3^{2} \cdot 5^{3}=18000$, so the number we are looking for can be 18000. (L.L.)