This preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: × 1 + 0 (So q 4 = 0). There±ore gcd(33 , 95) = 1. 5 Solve 33 x + 95 y = gcd(33 , 95) Method 1 We have that: 1 = 297 × 4 = 297(331 × 29) = 29 × (17(1)) + 33 × (7) = (952 × 33) × 8 + 33 × (7) = 95 × 8 + 33 × (2 × 87) = 95 × 8 + 33 × (23) So x = 8 and y =7. 6 Solve 33 x + 95 y = gcd(33 , 95) Method 2 To fnd x , we have that: x 2 =2 × x 1 + x =2 (as q 1 = 2) x 3 =1 × x 2 + x 1 = 3 (as q 2 = 1) x 4 =7 × x 3 + x 2 =23 (as q 3 = 7) To fnd y , we have that: y 2 =2 × y 1 + y = 1 y 3 =1 × y 2 + y 1 =1 y 4 =7 × y 3 + y 2 = 8 So x = 8 and y =7. 7...
View
Full Document
 Spring '10
 jhhgjhgjh
 Greatest common divisor, Euclidean algorithm, Articles with example pseudocode, Euclid, Extended Euclidean algorithm, Euclidean domain

Click to edit the document details