problem: What is the positive difference between the sum of the first 20 positive multiples of 5 and the sum of the first 20 positive, even integers?
solution: The sum of the first 20 positive multiples of 5 is $5+10+15+\cdots+95+100 = 5 (1 + 2 + \dots + 20)$.  For all $n$, $1 + 2 + \dots + n = n(n + 1)/2$, so $5 (1 + 2 + \dots + 20) = 5 \cdot 20 \cdot 21/2 = 1050$.
  
  The sum of the first 20 positive even integers is $2+4+6+8+\cdots+38+40 = 2 (1 + 2 + \dots + 20) = 2 \cdot 20 \cdot 21/2 = 420$.  The difference is $1050-420=\boxed{630}.$
  
  Alternatively, we can subtract the sums, to get
  \begin{align*}
  5(1 + 2 + \dots + 20) - 2(1 + 2 + \dots + 20) &= 3 (1 + 2 + \dots + 20) \\
  &= 3 \cdot \frac{20 \cdot 21}{2} \\
  &= 630.
  \end{align*}