0 votes 0 votes In a RSA cryptosystem a participant uses two prime numbers p and q is 17 and 11 respectively to generate his/her public and private keys...If the public keybof participant is 7 and cipher text C is 11 then the original message M is_________ dharmesh7 asked Jan 15, 2019 dharmesh7 443 views answer comment Share Follow See all 3 Comments See all 3 3 Comments reply dharmesh7 commented Jan 15, 2019 reply Follow Share @HeadShot 2 @Mk Utkarsh .how to solve the last step? Pls mention the source or tell the method...even i have applied Fermat's theorem..still i am mot able to solve it Caption 0 votes 0 votes Abhisek Tiwari 4 commented Jan 15, 2019 reply Follow Share https://gateoverflow.in/39588/gate2016-2-29 see laxman ans comments same one discussed. 0 votes 0 votes arya_stark commented Jan 15, 2019 reply Follow Share 11^2mod187 = 121 11^3mod187 = 22 then brk then in powers $11^{23}mod187 = ( (11^3mod187)^7 * 11^2mod187 )m od187$ $=> (22^7 * 121)mod187$ $=>(22^7mod187 * 121)mod187$ $=>(((22^2mod187)^3 * 22 )mod187 * 121 )mod187 $ $=> ((110^3*22)mod187 *121)mod187$ $=> (44*121)mod187$ $=> 5324mod187$ $=> 88$ 0 votes 0 votes Please log in or register to add a comment.