Question

1. Solve the following for x

   1. 4x ≡ 7 (mod 19)
   2. 6x ≡ 8 (mod 31)
   3. 9x ≡ 7 (mod 16)

Answer 1

42=7 (mad 14) ax= b[modn) ansa Stap Geoca,n) CACOCA, 19) = 1 Scaph b = 1 7 Someron pessile erists sap 3 d Cnodn) = (mod12) = 1 solution en saph. Divide beth Sudas by d of eat D sin mod 1a) Seaps: Multiplying both sides by multiplicative in way axu 'n = 7x4 (mod 19) = 7*4" (med) (4X c) mod n c=1 4 modia c=2 3 8 modig (3 3 12 modis l ti #1 & 1 a 125]. (352) 20 mod 19 = 1 Stape Replacing 4 by [635] ne 7x5 madia ar 36 Cod m aban31 gap! GED (6,31) = 1 dai Staph. d (moda) - Ilmod 31) Isomon exist Dividing both side by a g en we go an = 8 (nad 3) Mutapiging both sidy by multipliceste Seeps

Steps (6x c) mod31 = 1 ca 2 omoa31 1 c=2 = 12 mod31 & c=3 2 18 mod 31 t 1 cou de mode # 1 c=5 30 mod 31 1 l = 57 6257 2 342 mod 31 = 1 6 Stap 7 no=8x57 mod31 To=22 (Any)

To s an = 7 (modle) an = blmodn) arab =7 no 16 & Sep incola, 1a) = al doi Steph: b = I=7 souten possible Stap 3: d (modn) (med 16) = 1 Solution possible. Steph: Divide beth sides of ear by di- an = 7cmad 16) St o p - Sieu bg mycket mere axal=rxa n = 7x9 (mod (c) (madlo). 71 Step 6 (axc) mod 16 = 1 C4 med at c=2 7 18 medici = 2. l 225 225 m Step 7 n =7x25 (mod 16)