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