Q1

Q1. Solve the following questions: a. Given gcd (a, b) = 24, find gcd (a, b, 16). b. Given gcd (a, b, c) = 12, find gcd (a, b, c, 16) c. Find gcd (200, 180, and 450). d. Find gcd (200, 180, 450, 610). a) Given gcd of (a,b)24. we have to find gcd of (a,b,16). gcd of (a,b,16)= gcd of(24,16). 24=16×1+8, 16=8×2+0. gcd of (a,b,16)=8. b) Given gcd of ( a,b,c)=12 We have to find gcd of ( a,b,c,16). gcd of( a,b,c,16)= gcd of(12,16). 16=12×1+4, 12=4×3+0. gcd of ( a,b,c,16)=4. C) we have to find gcd of ( 200,180,450).

Eculiden Algo; q r1 r2 r 1 200 18 0 2 0 9 180 20 0 20 0 gcd(200,180) 20 q r1 r2 r 0 20 45 0 2 0 22 450 20 1 0 2 20 10 0 10 0 gcd(20,450) 10 gcd(200,180.4 50) 10 d. Find gcd (200, 180, 450, 610). Eculiden Algo; q r1 r2 r 1 200 18 0 2 0 9 180 20 0 20 0 gcd(200,180) 20 q r1 r2 r 0 20 45 2

The result of 0 mod 15 is 0 Q3. Let us assign numeric values to the uppercase alphabet (A = 0, B = 1, . . . Z = 25). We can now do modular arithmetic on the system using modulo 26. a. What is (A + N) mod 26 in this system? b. What is (A + 6) mod 26 in this system? c. What is (Y − 5) mod 26 in this system? d. What is (C −10) mod 26 in this system? A 0 K 10 U 20 B 1 L 11 V 21 C 2 M 12 W 22 D 3 N 13 X 23 E 4 O 14 Y 24 F 5 P 15 Z 25 G 6 Q 16 H 7 R 17 I 8 S 18 J 9 T 19

As per the given table - (a) (A+N) mod 26 Here, as per the table, A=0 and N=13 => 13 mod 26 = 13 (b) (A+6) mod 26 A=0 => 6 mod 26 = 6 (c) (Y-5) mod 26 Here, as per the table, Y=24 => (24-5) mod 26 = 19 mod 26 = 19 (d) (C-10) mod 26 Here, as per the table, C=2 => (2-10) mod 26 = (-8) mod 26 = 18

c. 9 x + 4 ≡ 12 (mod 7) → 9 x ≡ (−4 + 12) (mod 7) → 9 x ≡ 1 (mod 7) a = 9, b = 1, n = 7 → d = gcd ( a , n ) = 1 Since d divides b , there is only one solution. Reduction: 9 x ≡ 1 (mod 7) x 0 = (9−1 × 1) (mod 7) = 4 d. 232 x + 42 ≡ 248 (mod 50) → 232 x ≡ 206 (mod 50) a = 232, b = 206, n = 50 → d = gcd ( a , n ) = 2 Since d divides b , there are two solutions. Reduction: 116 x ≡ 103 (mod 25) → 16 x ≡ 3 (mod 25) x 0 = (16−1 × 3) (mod 25) = 8 x 1 = 8 + 50/2 = 33 Q5. List all additive inverse pairs in modulus 20. All additive inverse pairs in modulus 20 :- (0,0) ,

(1,19), (2,18), (3,17), (4,16), (5,15), (6,14), (7,13), (8,12), (9,11), (10,10).