A sequence of numbers $1, 4, 7, 10, \cdots, 697, 700$ follows the rule: the first number is 1, and each subsequent number is 3 more than the previous one, up to 700. If all these numbers are multiplied together, find the number of trailing zeros in the resulting product (for example, the number of trailing zeros in 12003000 is 3).
I. Solution

First, find the number of factors of 5 in the product.
In the array: $10, 25, 40, 55, \cdots, 700$ all contain the factor 5. This sequence has $(700-10) \div 15+1=47$ numbers.
If each number is counted with one 5, there are 47 fives in total;
However, among them: $25, 100, 175, \cdots 700$ also contain a second factor of 5, these numbers have
$$
\text { (700-25) } \div 75+1=10 \text { numbers. }
$$

Therefore, 10 more fives should be added.
Similarly, 250, 625 will have a third factor of 5, and 625 will have a fourth factor of 5.
Thus, the number of factors of 5 in the product is:
$$
47+10+2+1=60 \text { fives. }
$$

As for the number of factors of 2 in the product, it is clearly more than 60, so the product has a total of 60 zeros at the end.