Hi, Masters!

Can I have some questions?

1) Why is it that (the inverse of 3)^3 will be -33 * 3 * 11^2 * 3 = 121 * 3 = 21 * 3 = 63 mod 100?

I assume 3*33 = 99 so it is -1 mod 100. So -33 came from here. But I don't know where 3* 11^2 * 3 come from..

2) y^31 = 3 mod 100.

I just memorised it should be y = 3^31^-1 mod 100. And 31^-1 is -9, so the equation should be 3^-9 mod 100.

How can I calculate 3^-9 mod 100...?

I guess I am poor at calculating the inverse.

Please help me out masters.

It is really difficult to get any information about higher congruence like y^32...