2587392 mod 157656 = 64896, reducing to gcd(157656, 64896). Then 157656 mod 64896 = 27864, 64896 mod 27864 = 9168, 27864 mod 9168 = 360, 9168 mod 360 = 168, 360 mod 168 = 24 and 168 mod 24 = ...
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1789287 mod 237656 = 125695, reducing to gcd(237656, 125695). Then 237656 mod 125695 = 111961, then remainder 13734, then 2089, then 1200, 889, 311, 267, 44, 3, 2 and finally remainder 1 followed by 0, yielding GCD 1. Cbse Class 9 ...
81000 mod 51000 = 30000, giving gcd(51000, 30000). Then 51000 mod 30000 = 21000, 30000 mod 21000 = 9000, 21000 mod 9000 = 3000 and 9000 mod 3000 = 0. With zero reached, the greatest common divisor is 3000. Ganita Manjari ...
First, 825 mod 375 = 75, reducing to gcd(375, 75). Next, 375 mod 75 = 0, giving gcd(75, 0). Since the remainder reaches zero, the algorithm terminates to output 75 as the greatest common divisor. Cbse Class 9 Maths Ganita Manjari ...
We can terminate immediately upon finding the very first common divisor. Since we count down from min(m, n), that first common factor is guaranteed to be the greatest common divisor without further scanning. Class 9 Chapter 11 The World of Algorithms ...