
#1
Jul1009, 05:25 PM

P: 25

Struggling to put a number through this as I keep getting my original number as the encrypted number too.
A = 11 p = 3 q = 5 n = pq = 15 z = (p1)(q1) = 2*4 = 8 k = coprime of z = 7 So, A=11 n=15 z=8 (Public key) k=7(Public key) kj = 1 (mod z) 7j = 1 (mod 8) for which I am getting j = 9/7 (private key) Start of encryption... A^k = E (mod n) 11^7 = E (mod 15) 19487171/15 = 1299144.733...... 1299144 * 15 = 19487160 E = 19487171  19487160 = 11 (which is what I started with) Tried using the decrypting part anyway and got.... E^j = A (mod n) 11^(9/7) = A (mod 15) 21.8239547419283/15 = 1.45493031612855 1 * 15 = 15 21.8239547419283  15 = 6.8239547419283 (which obviously isnt what I started with) Any help where I am going wrong would be appreciated, I assume it is where mod is brought in as I havent used that function before 2 hours ago but it may be somewhere else. Cheers james 



#3
Jul1009, 05:38 PM

P: 25

I'm not sure. Thats where I think I have gone wrong though.
Think I messed up at kj = 1 (mod z) 7j = 1 (mod 8) trying to work out the private key as I have tried to teach myself how to do this from an example on a website without really knowing how to do it. James 



#4
Jul1209, 01:18 PM

P: 25

RSA Algorithm Help
Can anyone help with this? Will ask my math/physics teacher tomorrow but wouldnt mind having another go at it first.
Cheers James 



#5
Jul1209, 01:45 PM

Admin
P: 22,671

I am sure these should be all integers, but I have not played with RSA for several years.




#6
Jul1209, 02:25 PM

P: 25

I think you are right but I am getting fractions when the modulus function is introduced as I'm unsure how it works.
James 



#7
Jul1209, 03:12 PM

Admin
P: 22,671

mod never gives fractions, mod is about finding remainder.
10 mod 3 = 1 12 mod 5 = 2 and so on. 



#8
Jul1209, 03:41 PM

Math
Emeritus
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Thanks
PF Gold
P: 38,881




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