3. Given five numbers in sequence are $13,12,15,25,20$, the product of each pair of adjacent numbers yields four numbers, the product of each pair of adjacent numbers among these four numbers yields three numbers, the product of each pair of adjacent numbers among these three numbers yields two numbers, and the product of these two numbers yields one number. How many consecutive 0s can be counted from the units place to the left in this final number (refer to the figure)?



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3.【Solution】To determine the number of trailing zeros in the product of several integers, you can use this method: decompose each multiplier into prime factors, add up the multiplicities of 2, and also add up the multiplicities of 5. The smaller of these two sums is the number of trailing zeros in the product.

We can calculate the multiplicities of the prime factors 2 and 5 separately. For this purpose, we draw two diagrams (as follows): the numbers in the diagrams are filled in as follows: taking the multiplicity of 2 as an example, the first number 13 in the first row does not contain the factor 2, so we fill in 0 at this position; the second number 12 contains 2 factors of 2 (12 = 2 × 2 × 3), so we fill in 2 at this position, and so on. The numbers in the rows below are the sums of the two numbers above them (because the multiplicity of the factor 2 in the product is the sum of the multiplicities of the factor 2 in each multiplier).

In this way, we have marked the multiplicities of the prime factors 2 and 5 in the prime factorization of each number in the circles in the two diagrams. Specifically, the multiplicity of 2 in the prime factorization of the number at the bottom is 10, and the multiplicity of 5 is 15, so it should have 10 trailing zeros.

Answer: It can have 10 consecutive zeros.