Question

: Solve the equation 34.x = 21 mod 55 by finding the inverse of 34 in 255.

Answer 1

(Griven) Solving a linear congruence using Euclid's Algorithm. Given : 34x5 21 (mod 55) » we want to work out a multiplicative inverse (modulo 55). że. 34 v = 1 (mod 55) , x=3t med roo) inverse exist meare x3 giv(mod 55 ) Just. 34 x = 1 mod so 34 V = 1- 55W 34 V + 55W=1 55 = 1x3y + 21 34: 1x 217 13 21 = 1X13 +8 13:1X8+5 8 = 1 x 5 +3 5. 1x3 +2 2: |x 84 ). 21 = 55-34 13 = 34-21 8 = 21-13 5 = 13 -8 3 = 8-5 Q = 5-3 1= 3-2 1= 3-2-3-(5-3) - 2x3-5 c) : 8 ( 6-9) - S = ax 8 - 3 x 5 = 2X8 - 3 (13-8 - 2x 8-3X13 +8x3 = 5*8 -3x13 - 25*(21-13) - 3X13 : 5x 21 - 8 X13 = 5X21 - 8 (34-21) - 13% 21 -8134 +13 (55-34) - 8 X34 = 13455 - 21x34 => 1 : 13455 - 21*34 Wat mod 55 334 24*34 - HA V=-31, W=13, V--21= 34 en zss to al modos + 34x34 = 1 (mod (55)

344 14 34x २। (mod55) ) 34.34 x : 3५/21 (mods)) 34x34 - 714 (mddss) ) *: 46trmolss) :54 miss) | (AM) Check : 4024 54x34 - 1836 1836 - 21 = 1815 diurisile bass