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 3.1k 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 Chetan28kumar commented Oct 23, 2018 reply Follow Share 1123 mod 187 = 149 procedure to get this value 149 as above? (or yu r using calculator or any hit and trial method ) 0 votes 0 votes ank73811 commented Oct 23, 2018 reply Follow Share Calculator bro as you will be provided with that 0 votes 0 votes Chetan28kumar commented Oct 23, 2018 reply Follow Share by using calculator the answer comes wrong and it is from GATE 2016 0 votes 0 votes ank73811 commented Oct 23, 2018 reply Follow Share 3 aa raha ha 0 votes 0 votes Chetan28kumar commented Oct 23, 2018 reply Follow Share mera 12 aa rha h(by using calculator) ,,bt actual ans is 4. 0 votes 0 votes Magma commented Jan 27, 2019 reply Follow Share yeah By using virtual calculator answer is : 149 but if we use normal scientific calculator answer will be : 88 but my question is that..we unable to reduce it by applying totient formula :3 0 votes 0 votes Shubhgupta commented Jan 27, 2019 reply Follow Share if number is very big then VC will give the wrong result and here you can not apply totient method so solve it by splitting method. 0 votes 0 votes Magma commented Jan 27, 2019 reply Follow Share by splitting method. @Shubhgupta How ?? 0 votes 0 votes 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.