How many numbers are in the list 2008,2003,1998,...8,3?

How many numbers are in the list 2008,2003,1998,...8,3

arithmetic sequence: 

$$\small{\text{ $ a_1,\, a_2,\, a_3,\, a_4,\, a_5,\, a_6,\, \cdots ,\, a_n $}} \qquad \small{\text{ $\boxed{ a_n=a_1+(n-1)\cdot d } \qquad $ or $\qquad \boxed{ n=1+\dfrac{a_n-a_1}{d} } $ }}$$

If the distance between the numbers   $$\small{\text{$d = -5$}}$$ and $$\small{\text{$a_1=2008$}}$$  and $$\small{\text{$a_n=3$}}$$  then we find n:

$$\small{\text{$ \begin{array}{rcl} n &=& 1+\dfrac{a_n-a_1}{d} \\\\ n &=& 1 + \dfrac{3-2008}{(-5)}\\\\ n &=& 1 + \dfrac{-2005}{(-5)}\\\\ n &=& 1 + \dfrac{2005}{5}\\\\ n &=& 1 + 401\\\\ n &=& 402 \end{array} $}}$$

There are 402 numbers in the list.

5, or if you're being an a*s then its 6 (2,0,8,3,1,9)

(2008-3)/5+1=501+1=502