By how many zeros does the number 2012! end?
The integer 10 is not prime: therefore, we cannot directly apply Legendre's formula. By factoring 10 into prime factors, we realize that the largest exponent \( n \) such that \( 10^n \) divides 2012! is the smallest of the two numbers \( v_{2} \) (2012!) and \( v_{5} \) (2012!). Legendre's formula directly proves that it is \( v_{5} \) (2012!). It is:

$$
\left[\frac{2012}{5}\right]+\left[\frac{2012}{25}\right]+\left[\frac{2012}{125}\right]+\left[\frac{2012}{625}\right]+\left[\frac{2012}{3125}\right]+\ldots=402+80+16+3+0+\ldots=501
$$

Therefore, the number 2012! ends with 501 zeros.