problem: The sum of 5 consecutive even integers is 4 less than the sum of the first 8 consecutive odd counting numbers. What is the smallest of the even integers?
solution: The first 8 odd positive integers are 1, 3, $\dots$, 15.  The sum of an arithmetic series is equal to the average of the first and last term, multiplied by the number of terms, so their sum is $(1 + 15)/2 \cdot 8 = 64$.
  
  Let the 5 consecutive even integers be $a$, $a + 2$, $a + 4$, $a + 6$, and $a + 8$.  Their sum is $5a + 20$.  But this is also $64 - 4 = 60$, so $5a + 20 = 60$.  Solving for $a$, we find $a = \boxed{8}$.