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 ...
Discussion Forum Latest Questions
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 ...
The divisor list would be generated in decreasing order rather than increasing order. Consequently, when determining the greatest common divisor, we must report the leftmost element instead of the rightmost one. Class 9 Ganita Manjari Part 2 chapter 11 question answer
(i) [(6aᵧ) + (4aₛ) + (5aₖ) + (4aₚ) + (4a₉)] / 23. (ii) With 13 nylon and 10 feather shuttles, the expression is [(13aₙ) + (10a𝒻)] / 23. Yes, both methods yield the exact same overall average. Ganita Manjari part 2 ...