jmc

Let $a,b,c$ be the number of 30 cent items, 2 dollar items, and 3 dollar items that Janice bought, respectively. Since there are 30 items in all, $a+b+c=30$. The total cost is 3000 cents, so $30a+200b+300c=3000$, which can be rewritten as $$\begin{array}{rl}30a+(30b+170b)+(30c+270c)& =3000\\ \Rightarrow 30(a+b+c)+170b+270c& =3000.\end{array}$$ Substituting $a+b+c=30$ gives $$\begin{array}{rl}30\cdot 30+170b+270c& =3000\\ \Rightarrow 170b+270c& =2100\\ \Rightarrow 17b+27c& =210.\end{array}$$ Thus, $17b+27c$ is a multiple of 10. Since $17b+27c=10(b+2c)+7(b+c)$, $7(b+c)$ is also a multiple of 10. 10 cannot divide 7, so 10 divides $b+c$. Janice bought 30 items, so the reasonable values of $b+c$ are $0,10,20,30$. If $b+c=0$, then $17b+27c=0$, which isn't true. If $b+c=20$, then the least possible value of $17b+27c$ is $17\cdot 20=340$, which is also impossible. By the same reasoning $b+c=30$ is also impossible. We conclude that $b+c=10$, namely that $b=6$ and $c=4$ to satisfy $17b+27c=210$. Thus $a=30-(b+c)=\boxed{20}$.