4. (New Zealand Mathematics Competition) Divide the set of numbers $\{1,2, \cdots, 10\}$ into two groups such that the product of the numbers in the first group $p_{1}$ is divisible by the product of the numbers in the second group $p_{2}$. Find the minimum value of $\frac{p_{1}}{p_{2}}$.
4. Solution: Since 7 is a prime number and cannot be canceled out, it must be in the first group, and $\frac{p_{1}}{p_{2}} \geqslant 7$. When $p_{1}=3$ $\times 5 \times 6 \times 7 \times 8, p_{2}=1 \times 2 \times 4 \times 9 \times 10$, the minimum value 7 can be achieved.