Mathematics in Ancient and Medieval India c7.s5.196

c7.s5.196

Now, a₃=(11t+90)/(-7) is an integer for t=9 $₃=(99+90)/(-7)=189/(-7)=-27$ $b₃=(90t+67.11)/(-7)=(90× 9+737)/(-7)=1547/(-7)$ $=-221$ $c₃=(81-67)/(-7)=14/(-7)=-2$