Thread: RSA encryption

This problem is for my discrete math class - so I hope this is in the right section.

This problem is to be done using only the computing power of a pocket calculator. Write out all your work in step-by-step form showing all computations.

The Message M=9 is encrypted in the RSA system using p=11,q=13 and e =7 which is relatively prime to (p-1)(q-1)

a.)Encrypt M to get C , which is to be transmitted.

b.)find d, the inverse of e(mod(p-1)(q-1)).

c.)using d, recover the original message M.

Hope some1 can help me out


Cheers

M is the message you want to send.

The public key is (n, e) , n = p*q , 143 = 11*13
C is the coded message and it is coded by C = M^e mod n

C= 48 = 9^7 mod 143, to decode we need d which is the inverse of e

d*e is congruent to 1 mod (#'s relatively prime to n)
d*e is congruent to 1 mod ((p-1)*(q-1)
d*9 is congruent to 1 mod 120
d = 103

We can now decode the coded message C which is 48 by
M = C^d mod n
M = 48^103 mod 143
M = 9

I am not an expert in modular arithmetic so if you want to add something please do. I just joined this forum resently and only saw your post a couple of days ago. I hope you got an answer long before today