How many digits.....

How many zeros are at the end of the whole number that is the product of the first 2012 positive integers?

There would be 3 zeros appended for each multiple of 1000.
There would be 2 zeros appended for each multiple of 100 that is not a multiple of 1000.
There would be 1 zero appended for each multiple of 10 that is not a multiple of 100.
There would be 1 zero appended for each multiple of 5 times an even number. There are less multiples of 5, so we will count them and multiply each with one of the excess of even numbers.

Between 1 and 2012, there are 2 multiples of 1000: 1000 and 2000
Between 1 and 2012, there are 18 multiples of 100 (not already counted): 100, 200, ... 900, 1100, ... 1900
Between 1 and 2012, there are 181 multiples of 10 (not already counted): 10, 20, ..., 90, 110, ... 190, 210, ... 1990, 2010
Between 1 and 2012, there are 201 multiples of 5 (not already counted): 5, 15, 25, ... 2005

Zeros = 2*3 + 18*2 + 181*1 + 201*1 = 6 + 36 + 181 + 201 = 424 zeros at the end of the number.

A zero occurs for every pair of 2 and 5 that you have. There will obviously be more 2's than 5's because 2's happen more often.
So there are 2012/5 or 402 5's (there cannot be part of a 5)
But you cannot forget about the numbers that have a second factor of 5(such as 25)
and there are 402/5 or 80 5's
then
80/5 = 16
16/5 = 3

so there are 402+80+16+3 5's or 501 5's, thus there will be 501 zeroes.

To be honest I don't know if this is correct because I did it in like 2 minutes, and there might be more to the problem I didn't catch..

Look here to see how to approach this problem.

http://nrich.maths.org/608/solution

You basically need to find all the multiples of 5 between 1 and 2012, taking into account that some have more than one factor of 5. For example 25 =5*5 has two 5's, 125=5*5*5 has 3 fives, 1000=5*5*5*8 has 3 fives, etc.

Count up all of the 5's and you will have the number of zeros at the end.

1 zero.