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# RSA Cipher

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Alice chooses two large prime numbers p and q. She finds their product, m=pq which is public. She also finds n= (p-1)(q-1) which is private. She chooses e which is a number relatively prime to n and finds d= the inverse of e (mod n). The number e is public and the number d is private. When Bob wants to send a message x (a number which represents a block of letters) to Alice, he calculates y=x^e (mod m) and sends it to Alice. She recieves y and finds the plan text x by computing x=y^d (mod m).

Suppose that p=7 and q=11
What is M?
What is n?
Suppose that e=7, what is d? (check by multiplying d and e together and reduce).
Suppose Alice recieves ciphertext y=51, what is the plaintext x?

##### Solution Summary

An RSA cipher is solved.

##### Solution Preview

p=7, q=11

1. m=pq=7*11=77
2. n=(p-1)(q-1)=6*10=60
3. e=7, then d=e^(-1)(mod n)=7^(-1) (mod 60). We note ...

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