$2013!$ ends in a string of zeros. How many of them are there?
Work
The answer is given like this:(Kindly explain why this is so)
$2$’s and $5$’s pair off to produce multiples of 10 and since there are more 2’s than 5’s in the prime factorisation of $2013!$ we need to find the number of 5’s in this factorisation.
$\frac{2013}5=402$ numbers are multiples of 5
$\frac{2013}{25} =80$ numbers are multiples of $5^2$
$\frac {2013} {5^3}=16$ numbers are multiples of $5^ 3$
$\frac {2013}{5^4}=3$ numbers are multiples of $5^4$
$\mbox{402+80+16+3 = 501}$ numbers are there in total so there will be 501 zeroes.
If you factor $2013!$ and take the exponent on $5$, that's your answer. For example: $$ 5! = 120 = 5^1 3^1 2^3 \Rightarrow 1\;\text{zero}\\ 6! = 720 = 5^1\dots \Rightarrow 1\;\text{zero}\\ 10! = 3,628,800 = \dots 5^2\dots \Rightarrow 2\;\text{zeros}\\ 15! = 1,307,674,368,000 = \dots 5^3\dots \Rightarrow 3\;\text{zeros} $$ But notice that when you get to $25$, this number will contribute to $+2$ on the exponent of $5$: $$ 20! = 20\dots 15\dots 10\dots 5\dots = \dots 5^4\dots \Rightarrow 4\;\text{zeros}\\ 25! = 25\dots 20\dots 15\dots 10\dots 5\dots = \dots 5^6\dots \Rightarrow 6\;\text{zeros} $$ That's why you count the multiples of $5$, then add $+1$ for each multiple of $25$, and so on for $125$, etc.