Class 9 Maths Ganita Manjari Chapter 11 Exercise 11.3 Solutions The World of Algorithms
The Class 9 Maths Ganita Manjari Chapter 11 Exercise 11.3 Solutions – The World of Algorithms help students understand how algorithms can be used to find divisors, determine the greatest common divisor (GCD), and work with prime numbers and prime factorisation. The exercise develops logical thinking by showing how mathematical problems can be broken into clear, step-by-step procedures.
The solutions are explained in a simple and systematic way so that students can understand the reasoning behind each algorithm and apply it correctly. Students can also access the Class 9 Maths Ganita Manjari Chapter 11 Exercise 11.3 Solutions PDF for convenient revision and offline study.
NCERT Class 9 Maths Ganita Manjari Chapter 11 Exercise Set 11.3
1. How would our original algorithm change if we computed the divisors of n by examining the numbers from 1 to n in reverse order; from n down to 1?
Answer:
If we examine the numbers from n down to 1, we will still find all the divisors of n, but they will be listed in decreasing order.
For example, the divisors of 18 would be:
18, 9, 6, 3, 2, 1
Thus, the algorithm for finding the divisors remains the same except that we check the numbers in reverse order.
If these divisor lists are then used to find the GCD of two numbers, the first common divisor we encounter will be the GCD, because the divisors are arranged from greatest to smallest.
So, the main change is:
- divisors are obtained in decreasing order; and
- for finding the GCD, we take the first common divisor instead of the last common divisor.
2. What about the last algorithm described above? What happens when we look at common divisors starting from min(m, n) and work backwards to 1?
Answer:
If we start checking numbers from min(m, n) and work backwards to 1, the first number that divides both m and n will be their GCD.
This is because min(m, n) is the largest possible value that a common divisor can have. Therefore, as we move downwards, the first common divisor we find must be the greatest common divisor.
So, the algorithm can stop as soon as the first common divisor is found.
For example, let:
m = 18 and n = 24
We start from min(18, 24) = 18 and check:
18, 17, 16, …, 6
The first number that divides both 18 and 24 is 6.
Therefore:
gcd(18, 24) = 6
This method can be more efficient because once the first common divisor is found, there is no need to check any smaller numbers.
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