Ask Your Question

how to solve this equation

asked 2014-11-07 07:33:42 -0500

Tummala Dhanvi gravatar image

(83359654581036155008716649031639683153293510843035531 ^x) % 32003240081 =1

edit retag flag offensive close merge delete

1 answer

Sort by ยป oldest newest most voted

answered 2014-11-07 08:36:58 -0500

kcrisman gravatar image

Here's a simpler example, solving $53^x \equiv 1\text{ mod }(71)$.

sage: a = mod(53, 71)
sage: one = mod(1,71)
sage: one.log(a)

In retrospect, this is not so surprising (though presumably not the answer you wanted). Indeed, part of the point of this exercise is probably to show that it's hard to solve such things by brute force.

For small examples you can...

sage: for i in range(1,70):
    if a^i==1:
        print i
<no output>

but I nearly crashed my computer trying to do yours that way, just as an experiment! So it's best to use some math. In this case, 53 was a primitive root of 71 - though your modulus is not prime. (Hint, hint.)

edit flag offensive delete link more

Your Answer

Please start posting anonymously - your entry will be published after you log in or create a new account.

Add Answer

Question Tools

1 follower


Asked: 2014-11-07 07:33:42 -0500

Seen: 419 times

Last updated: Nov 07 '14