imc

Let $x$ be the number of $5$-cent coins that Joe has. Therefore, he must have $(x+3)10$-cent coins and $(23-(x+3)-x)25$-cent coins. Since the total value of his collection is $320$ cents, we can write $$\begin{array}{rl}5x+10(x+3)+25(23-(x+3)-x)& =320\\ 5x+10x+30+500-50x& =320\\ 35x& =210\\ x& =6.\end{array}$$ Joe has six $5$-cent coins, nine $10$-cent coins, and eight $25$-cent coins. Thus, our answer is $8-6=\boxed{2}.$