Question

6.) Find (i) 225 mod 21, (ii) 766 mod 120 and (iii) the last two digits of 1 + 7162 + 5121. 3.12.

Answer 1

20 212 solution page no. © 225 modal Now gcd (2,21) = 1 do By Eulers theorem ORI) El modzi 2 21- 783 $(21) = 6X2 = 12 12 2 El modzi = 1.1 modzi . If a = b modc a² = 6² mode 224 = 1 modal =) G 24 2.2 2.1 modas If a = b mode & KE ZI ka = kob mode = 2 modal 25 2

Now 1 ③ modizo page 2 gel (7, 120) = By Euters Theorem 00120) = 1 mod 120 그 Now 3 (20= 2x3x5 3 $ (120) = 4X2 X 4 = 32 32 T = 1 mod izo 7. 73.1 mod 120 con If a = b moda a mode 64 =) =1 mod 120 Is 3) 64 7. 7 = 7. ?? modizo to If a=bmode & kez > Kia= krb moda 66 73 72 mod 120 1 = 49 mod 120

page ③ ③ Least two digits of 1+ 7162 512 3312 is Itlast two digits of 7162) + (last two digits of 5121, 3322 ) Now first we two digits of are going to find last 517 3312 Note gcd (5,100) 1 But for any n7,2 5h has 25 as last two digits 12) are 25 (Р Now so last two digits of 5 ged (3,100) = 1 = By Eulers theorem 3 betoo) - I mod loo 10o = 5² x 2² o ciao) = 5(4) 2 (1) = 20x2 6° ° coo) = 40 40 3 E I mod coo (940)=(1?mod loro (If a = b modc 4 non > an= b mode) 280 = 1 modloo Now, 32 _ 1853020188 85 1841 32 = 41 mod loo 32

page 6 & from ® 4 6 280 3 32 3 1 41 modroo 812 3 3 h 41 modico then are 41 (o If aab mode of d = f mode ad bf mode) last two digits of 9812 s from @4@ we conclude that last two digits of 3312 are last two digits of 25.41 5t24 Now 25-41 = 1025 = last two digits of 25.41 are 25 o last two digits of sy 3312 25 Now gce (7, loo) = 1 dcloo) 7 E I modloo El modico (4 = cu4 mod cico) (o. If arb mode 4 nG N then an= 5n mode) 1 modloo 4 modroo = 71625 ug mod loo = last two digits of 7162 are 49 160 7 2 49 2

page ③ o last two two digits of 1 + t 5121 35 162 312 = 1+ 49 +25 - 75 312 o last two digits of 1 + 7/24 s'?! ₃ are 75