4.4 How many positive integer factors does the number 20! have?
(Mexico Mathematical Olympiad, 1990)
[Solution] Since the prime numbers less than 20 are $2,3,5,7,11,13,17,19$.
In the standard factorization of 20!
the highest power of 2
$$
\begin{array}{l}
=\left[\frac{20}{2}\right]+\left[\frac{20}{4}\right]+\left[\frac{20}{8}\right]+\left[\frac{20}{16}\right] \\
=18 .
\end{array}
$$
the highest power of 3
$$
\begin{array}{l}
=\left[\frac{20}{3}\right]+\left[\frac{20}{9}\right] \\
=8 .
\end{array}
$$
the highest powers of $5,7,11,13,17,19$ are $4,2,1,1,1,1$ respectively.
Thus, the standard factorization of 20! is
$$
20!=2^{18} \cdot 3^{8} \cdot 5^{4} \cdot 7^{2} \cdot 11 \cdot 13 \cdot 17 \cdot 19 .
$$

Therefore, the number of positive divisors of 20! $r(20!)$ is
$$
\begin{aligned}
& \tau(20!) \\
= & (18+1)(8+1)(4+1)(2+1)(1+1)(1+1)(1+1)(1+1) \\
= & 19 \cdot 9 \cdot 5 \cdot 3 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \\
= & 41040 .
\end{aligned}
$$