How many consecutive zeroes end 2000!, exactly?

Answer 1

Each zero at the end requires a 5 and a 2 somewhere in the factors. Since there are more twos than 5s, all you need to do is count the number of 5s. That's still an awfully big task. Each number that is divisible by 5 has at least one 5 as a factor (e.g., 5, 10, 15...) Of course there are 2000/5 = 400 of these.

Each number divisible by 25 has at least 2 factors of 5. There are 2000/25 of these or 80. We already accounted for one of these 5s in the first count so this adds 80 more.

Each number divisible by 5^3 (125) add another factor of 5 for 2000/125 = 16.

Now, on to 5^4 this time the division is uneven: 2000/625 = 3.2 This means there are 3 numbers below 2000 divisible by 5^4

Once we get to 5^5 = 3125, it is bigger than 2000 so there are no more factors of 5.

All in all this comes to: 400 + 80 + 16 + 3 = 499 zeros.

You also asked about 1000!. Using the same approach:

200 + 40 + 8 + 1 = 249 zeros

Answer 2

lets see one per every ten, including 1 for absulot zero, 201 zeros could be counted, 10 has a zero, 2000/10=200+1 for absalout zero. 201 zeros in 2000.