2017!

When multiplied out, the number 2017!=1x2x3....x2015x2016x2017 end with a striking of zeros.

How many zeros are there at the end of this number.

Let n = 2017

Legendre's Theorem - The Prime Factorization of Factorials:

$\begin{array}{|r|r|c|l|} \hline \text{p=factor} & \text{to base}_p & s=\text{ the sum of all the digits} & \text{exponent }= \frac{n-s} {\text{p }-1} \\ & & \text{in the expansion of n in base p } & \\ \hline 2 & 11111100001_2 & 7 & \frac{2017-7}{2-1} = 2010 \\ 3 & 2202201_3 & 9 & \frac{2017-9}{3-1} = 1004 \\ 5 & 31032_5 & 9 & \frac{2017-9}{5-1} = 502 \\ \cdots & \cdots & \cdots & \cdots \\ \hline \end{array}$

Prime factorization on 2017!:

$2^{2010}\times 3^{1004}\times 5^{502}\times 7^{334}\times 11^{200}\times \ldots \times 1999\times 2003\times 2011\times 2017$

Zeros at the end of 2017!:

$\begin{array}{|rcll|} \hline 2^{502} \times 5^{502} &=& (2\cdot5)^{502}=10^{502} \\ \hline \end{array}$

There are 502 zeros at the end of 2017!

2017 / 5 =403

2017/25 =80

2017/125 =16

2017/625 =3

Number of trailing zeros of 2017! =403 + 80 + 16 + 3 =502.