$1 \cdot 12$ 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).
(China Junior High School Mathematics League, 1988)
[Solution] First, find the number of factor 5s in the product.
In the array, $10, 25, 40, 55, \cdots, 700$ all contain the factor 5. This sequence has
$$
(700-10) \div 15+1=47 \text{ numbers. }
$$

This means that each number contributes one 5, totaling 47 fives.
However, among them, $25, 100, 175, \cdots, 700$ contain a second factor 5, and these numbers total
$$
(700-25) \div 75+1=10 \text{ numbers. }
$$

Therefore, we should add another 10 fives.
Similarly, 250 and 625 contain a third factor 5, and 625 contains a fourth factor 5. Thus, the number of factor 5s in the product is
$$
47+10+2+1=60 \text{ fives. }
$$

Since the number of factor 2s in the product is more than 60, the product ends with 60 consecutive zeros.