Let's analyze: the product that can get 0 must be

(1) 1. A number ending in 0 is set to k.

From 1 to 199 1, the last digit is zero.

10 20 30 40…… 1990

From 1 to 20, there are two such numbers (ending with zero).

There are three such numbers from 1 to 30.

From 1 to 10, there are 1 1.

1 120 12

Then from 1 to 199 1, there are1990/10 =199 bits with zeros.

That is, k has 199 values.

2. In addition, there are two zeros at the end, including1900/100 =19 (similar).

In the previous step, there were two zeros at the end of each number, and one zero was omitted.

Add 19

3. If there are three zeros at the end, there is 1000, and two zeros are missing.

Plus 2

(2) The corresponding last digit is the product of 2 and 5:

Add 2 and 5 to the corresponding value of each k to get the required number.

However, it is found that 2 and 5 are not included in this calculation, and 1990 plus 2 5 is no longer in the range, and one more and one less is exactly equal to the number of k 199.

So the answer is199+199+19+2 = 419 zeros.

Sixth floor, man. You almost said it.