Example 1 If a positive integer has 21 positive divisors, what is the smallest such positive integer?
Analysis We use the unique factorization theorem to handle this.


First, let's set the prime factorization of the positive integer \( N \) as \( N = p_{1}^{t} p_{i}^{t} \cdots p_{k}^{t} \).
Here, \( l_{1} \geqslant 1 \) (for \( i=1,2, \cdots, k \)), and \( p_{1}, p_{2}, \cdots, p_{k} \) are distinct prime numbers. We will calculate the number of positive divisors of \( N \). Clearly, the positive divisors of \( N \) are \( d = p_{1}^{m_{1}} p_{2}^{m_{2}} \cdots p_{k}^{m_{k}} \), where \( 0 \leqslant m_{1} \leqslant l_{1}, 0 \leqslant m_{2} \leqslant l_{2}, \cdots, 0 \leqslant m_{k} \leqslant l_{k} \). When \( m_{i} \) takes all values from \( 0, 1, 2, \cdots, l_{i} \), \( d \) takes all positive divisors of \( N \). Therefore, the number of positive divisors of \( N \) is
\[
\prod_{i=1}^{k}\left(l_{i}+1\right) = \left(l_{1}+1\right)\left(l_{2}+1\right) \cdots \left(l_{k}+1\right).
\]

Second, suppose the positive integer \( N \) has 21 positive divisors. If we still set \( N = p_{1}^{l_{1}} p_{2}^{l_{2}} \cdots p_{k}^{l_{k}} \), then we have
\[
\left(l_{1}+1\right)\left(l_{2}+1\right) \cdots \left(l_{k}+1\right) = 21 = 3 \times 7 \times 1 \times \cdots \times 1.
\]

Thus, \( l_{1}+1=3, l_{2}+1=7, l_{3}+1=1, \cdots, l_{k}+1=1 \).
This gives \( l_{1}=2, l_{2}=6, l_{3}=\cdots=l_{k}=0 \).
Therefore, a positive integer \( N \) with 21 positive divisors must have the form: \( N = p_{1}^{2} \cdot p_{2}^{6} \).
If we take \( p_{1}=3 \) and \( p_{2}=2 \), then \( N = 2^{6} \cdot 3^{2} = 576 \) is the smallest positive integer that satisfies the condition.