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A big number modulo a prime 
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#1
Dec1212, 09:06 AM

P: 26

Im looking through old exams for a course in Cryptography and have beaten my head against the wall for a long time on one of the questions:
p = 683 is a prime, p1 = 2*11*31. What is x = 4^11112 mod p? When i did chinese remainder theorem on primes 2,11,31 i got that x = 16 mod 682, but so far i have not found a way to use this in determining x mod 683.. Also, when computing discrete logs I have found that one goes from mod p to mod (p1) alot, why is that? 


#2
Dec1212, 10:01 AM

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P: 11,925

Can you use a computer to calculate it?
The common general approach to calculate a^b mod c will give you the answer in milliseconds. 


#3
Dec1212, 10:14 AM

P: 26

It is supposedly solvable by hand, but I'm wondering if it was meant to calculate mod p1 instead of p. Atleast thats what I'll conclude if no one finds another answer.
Also, it seems that the answer is 16 mod 683 as well. Perhaps there is some onceinalifetime connection between 682 and 683 for this task? 


#4
Dec1212, 10:19 AM

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P: 11,925

A big number modulo a prime
4^11 = 2^22 = 1 mod 683
I would expect that there is some way to get this from p1=2*11*31. The rest is easy, as 11112 = 11*1010 + 2. 4^6 = 2 mod 683 is quite interesting, too. It is possible to calculate that by hand, but then you don't need the hint. 


#5
Dec1212, 11:15 AM

P: 26

4 is quadratic residue, so 4^(p1)/2 = 1 mod p, but then i get 4^11*31 = 1 mod p, and i need to get rid of 31.. Otherwise i get stuck with 4^200 mod p. I'll try to figure something out, thanks for the help!



#6
Dec1212, 11:22 AM

P: 894




#7
Dec1212, 11:52 AM

P: 26

Fermat gives 4^11112 = 1*4^200 mod p, so that gets me a little further, but not quite there.



#8
Dec1212, 12:19 PM

P: 894




#9
Dec1212, 12:29 PM

P: 26

That does help, its just that i want to find a better way of doing it! Usually these things turn out to be some number squared or something, so I feel like im doing something wrong if i compute too much :p



#10
Dec1212, 12:32 PM

P: 688

p1=2*11*31 is also phi(p), and the order of any number (for example, 2) mod p is going to be a divisor of phi(p). So there are not many combinations to try in order to find a divisor d of phi(p) such that 2^d =1 mod p.



#11
Dec1212, 02:08 PM

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P: 11,925




#12
Dec1312, 05:19 AM

P: 26

Indeed, that does help. Thank you everyone for helping out!



#13
Dec1512, 02:55 PM

P: 359

The way I did it was
4^682=1 mod 683 (4^219)(4^463)=1mod 683 Then 4^11=1 mod 683 You can surely simplify 4^219 by using that The rest should be clear. 


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