Let $x_1$, $x_2$, $x_5$, $x_{10}$, $x_{20}$, $x_{50}$ and $x_{100}$ be the ammounts of 1, 2, 5, 10, 20, 50 and 100 paises when we give $A$ paises with $B$ coins. Then $$A=x_1+2x_2+5x_5+10x_{10}+20x_{20}+50x_{50}+100x_{100}$$ and $$B=x_1+x_2+x_5+x_{10}+x_{20}+x_{50}+x_{100}.$$
Then $$100B=100x_1+100x_2+100x_5+100x_{10}+100x_{20}+100x_{50}+100x_{100}=\\\=100(x_1)+50(2x_2)+20(5x_5)+10(10x_{10})+5(20x_{20})+2(50x_{50})+(100x_{100}).$$
So we can give $100x_{100}$ 1 paise coins, $50x_{50}$ 2 paise coins and so on.