200 1 will not generate 0 by 2004, so it will be discussed in 2000.

Method 1:

The product will produce 0, that is, a multiple of 2 times a multiple of 5, such as 8× 15= 120, and so on.

From 1 to 3000, the multiple of 2 is greater than the multiple of 5, so just find out what the factor of 5 is and how many zeros there are in the end.

2000÷5=400

400÷5=80

80÷5= 16

16÷5=3.2, and the integer is 3;

So the factor of 5 is: 400+80+ 16+3=499.

therefore

The number of zeros at the end is: 499.

Method 2:

Because: 10=2×5

There are m zeros at the end of the product, that is, there are m factors 5, multiplied by m factors 2. Because in 1-2000, factor 2 is obviously more than factor 5, so we only need to figure out how many factors 5 are:

(1) Use four factors 5625 = 5×5×5×5,

2000÷625=3.2, and there are 3 in 1-2000;

② If there are three factors 5, 125=5×5×5,

125 =16 in 2000, and1-0/6 in 2000;

③ There are two factors: 5,25 = 5× 5,2000÷25 = 80,1-80 in 2000;

(4) 1 factor 5, 5 =1× 5,2000 ÷ 5 = 400, 1-2000 has 400;

Total * * *:

400+80+ 16+3=499