jmc

If the two people received one fewer coin, then the number of gold coins you would have would be a multiple of 11. However, there are two extra coins there, so the number of gold coins you have can be written in the form $11k+2$. We have that $11k+2<100$, so $k<\frac{98}{11}$. Since $k$ is the number of gold coins each person is receiving, $k$ must be an integer, so we have that $k=8$. Therefore, the largest number of gold coins you could have is $11(8)+2=\boxed{90}$.