1 votes 1 votes In a RSA cryptosystem , a participant uses two prime numbers p and Q is 17 and 11 respectively to generate his/her public key and private keys. if the public key of participant is 7 and cipher text(C) is 11, then the original message (M) is______? Chetan28kumar asked Oct 23, 2018 Chetan28kumar 3.0k views answer comment Share Follow See all 2 Comments See all 2 2 Comments reply Shubhanshu commented Oct 23, 2018 reply Follow Share 149?? 0 votes 0 votes Chetan28kumar commented Oct 23, 2018 reply Follow Share given ans is as above .bt how this one will came! .may yu explain? 0 votes 0 votes Please log in or register to add a comment.
0 votes 0 votes Public Key = (n, e) e = 7 , p = 17 , q = 11 Private Key = (n, d) n = pq = 187 Φ(n) = (p-1)(q-1) so, Φ(n) = 160 d *e mod Φ(n) = 1 d = 23 cipher text , c = 11 Message , M = cd mod n M = 1123 mod 187 = 149 ank73811 answered Oct 23, 2018 ank73811 comment Share Follow See all 11 Comments See all 11 11 Comments reply Show 8 previous comments Shubhgupta commented Jan 27, 2019 reply Follow Share $11^{23}mod187 =(11^{10}∗11^{10}∗11^{3})mod187 =(11^{10}mod187∗11^{10}mod187∗11^{3}mod187)mod187 =(66∗66∗22)mod187 =88$ split big power to small small power then VC will show the correct result. 2 votes 2 votes Magma commented Jan 27, 2019 reply Follow Share @Shubhgupta thanks 0 votes 0 votes mehul vaidya commented Feb 12, 2019 reply Follow Share can someone tell me how 23 calculated . I mean how actual calculation is done? 0 votes 0 votes Please log in or register to add a comment.
0 votes 0 votes In RSA we encrypt using public and decrypt using private key phi = (p-1)(q-1) = 160 Public key = 7 calculate Private key(d) comes out to be 23 let original msg be 'm' ciphertext be 'c' c^d mod n = m // as c = m^e mod n and by euler toitent we know m^ed mod n = m c = 11, d = 23 and n = p*q calculate m ! adityaaswal answered Oct 24, 2018 • edited Jan 27, 2019 by adityaaswal adityaaswal comment Share Follow See all 0 reply Please log in or register to add a comment.