How many numbers are in the list $2008, 2003, 1998, \ldots, 8, 3?$
I see that this is an arithmetic sequence, where each term is 5 less than the previous term. I want to find out how many terms there are in the sequence, so I need to find a formula for the nth term in terms of the first term and the common difference. I recall that the formula for an arithmetic sequence is $a_n = a_1 + (n-1)d$, where $a_n$ is the nth term, $a_1$ is the first term, and $d$ is the common difference. In this case, $a_1 = 2008$, $d = -5$, and $a_n = 3$, since 3 is the last term in the list. I plug these values into the formula and get $3 = 2008 + (n-1)(-5)$. I simplify and solve for $n$: $-2005 = (n-1)(-5)$, $401 = n-1$, $n = 402$. Therefore, there are 402 numbers in the list. 