Originally Posted by

**chrisc** __Given that:__

3^6 = 44 (mod 137)

3^10 = 2 (mod 137)

find x such that 3^x = 11 (mod 137)

such that 0 < x < 135

__My attempt:__

I first noticed that fact that 11 = (44/2)/2

So I tried using that.

I though maybe I can subtract the exponents and test my luck.

(3^6)/(3^10) = (3^-4)

(3^-4)/(3^10) = (3^-14)

-14 + 137 = 123

Doesn't work out. And Im sure there are a few things wrong with this.

Ive been looking at this question for quite some time. So now I am here for advice.

The answer I am looking for is **122**.