Geesshhhh...did you know that 9 mod 23 = 9?
11^a = 9
a = LOG(9) / LOG(11)
Take over, Rover :cool:
What is 30 mod 23?
Are you studying something like what they discuss here? Also, are you working with "discrete" logs?11^a=9 (mod 23)
I know that we have to use logarithms here, but do not know how to implement it... Could someone explain me how to solve it. Thank you in advance.
Are you studying something like what they discuss here? Also, are you working with "discrete" logs?
Thank you!
You are right man) I have searched for it in the internet, and I found couple algorithms that are based on trivial mutiplication. My teacher told me, if I solve this problem, I do not need to participate in final exam. Now I understand why she said like that.