0 votes 0 votes If p is any prime number such that a p-1 =r (mod p) and p=3 , a=6 then value of r is ________________ Computer Networks computer-networks network-security ace-test-series + – Agam asked Jan 30, 2018 • edited Mar 3, 2019 by I_am_winner Agam 461 views answer comment Share Follow See all 4 Comments See all 4 4 Comments reply Rishabh Gupta 2 commented Jan 30, 2018 reply Follow Share 0. Just put the values and solve it. These are so small values. 0 votes 0 votes Shubhanshu commented Jan 30, 2018 reply Follow Share You might took it as Fermate Little Theorem example:) It is fallacy to Fermat's Little Theorem. For applicability of FLT i.e. $a^{p-1} = 1 (mod p)$ p should not be divide a. But here p = 3 divides a=6. Hence it is a Fallacy to FLT. 0 votes 0 votes hs_yadav commented Jan 30, 2018 reply Follow Share use fermet theorem... (3*2)3-1 / 3 ....(3)3-1 / 3 * (2)3-1 / 3 => 0*1=0 0 votes 0 votes divyan commented Jan 1, 2019 reply Follow Share It is being true for all 0,3,6,9,.... As a^(P-1) = r(mod P) Now putting values 6^2 = r(mod P) 36-r = 0(mod 3) Now if you will put r = 0 or 3 or 6 or 9 and so on... It will satisfy. Are we supposed to choose minimum no.? 0 votes 0 votes Please log in or register to add a comment.