Question

In an RSA system, the public key of a given user is e = 31, n = 3599. What is the private key of this user? Hint: First use trial-and-error to determine p and q; then use the extended Euclidean algorithm to find the multiplicative inverse of 31 modulo \pi (n)

Answer 1

Answer:

p.followme

Explanation:

p=5

q=11

n=pq=55

t=(p−1)(q−1)=(5−1)(11−1)=40

e is chosen so as to be coprime to t=40 (and you picked e=7 Wow! – that was lucky It’s almost like you knew!)

Now pick a d such that de≡1 mod 40

d=23 (Because de=23×7=161=4×40+1)

d is your Private Key exponent, which you keep a dark, dark secret.

e is your Public Key Exponent, which gets published along with the value of n

Now anyone who wants to send you a plaintext message M, can encrypt it by using the Public Key (e,n)

X=Me mod n=M7 mod 55

and the cipher can be decrypted by using the Private Key (d,n)

M=Xd mod n=X23 mod 55