Number Play Class 6 Questions and Answers - Free PDF Download
Chapter 3 Number Play of Class 6 Maths introduces students to the fascinating world of numbers through various concepts like prime numbers, divisibility, factors, multiples, and magic numbers. Exercise 3.6 focuses on the Magic Number of Kaprekar, a unique mathematical discovery by the Indian mathematician D. R. Kaprekar. This exercise helps students explore how specific numbers can behave in interesting ways and strengthens their understanding of number operations and patterns.
Table of Content
Our Class 6 Maths NCERT Solutions PDF breaks the lesson into easy-to-understand explanations, making learning fun and interactive. Students will develop essential language skills with engaging activities and exercises. Check out the revised CBSE Class 6 Maths Syllabus and start practising the Maths Class 6 Chapter 3.
Glance on Class 6 Chapter 3 Exercise 3.6 The Magic Number of Kaprekar
Introduction to Kaprekar's Magic Number (6174).
Understanding how subtracting a number from its reverse leads to the magic number.
Exploration of digit manipulation and their properties.
Mental calculations and number puzzles.
Prime numbers, composite numbers, and their uses in number play.
FAQs on NCERT Solutions For Class 6 Maths Chapter 3 Number Play Exercise 3.6 - 2025-26
1. How do I find the Highest Common Factor (HCF) of 18 and 48 using the prime factorisation method as per NCERT Class 6 Maths Chapter 3?
To find the HCF of 18 and 48 using prime factorisation, you follow these steps:
- Step 1: Find the prime factors of the first number. For 18, the prime factorisation is 2 × 3 × 3.
- Step 2: Find the prime factors of the second number. For 48, the prime factorisation is 2 × 2 × 2 × 2 × 3.
- Step 3: Identify the common prime factors in both lists. Here, both numbers share one '2' and one '3'.
- Step 4: Multiply these common prime factors together to get the HCF. So, HCF = 2 × 3 = 6.
For more detailed methods, you can refer to Vedantu's page on HCF (Highest Common Factor).
2. What is the step-by-step process to find the Lowest Common Multiple (LCM) of 20, 25, and 30 as shown in Chapter 3 solutions?
The correct method to find the LCM of 20, 25, and 30 using prime factorisation is as follows:
- Step 1: Write the prime factorisation for each number.
20 = 2 × 2 × 5 = 2² × 5
25 = 5 × 5 = 5²
30 = 2 × 3 × 5 - Step 2: Identify the highest power of each prime factor that appears in any of the factorisations. The prime factors are 2, 3, and 5. The highest power of 2 is 2², the highest power of 3 is 3¹, and the highest power of 5 is 5².
- Step 3: Multiply these highest powers together to find the LCM.
LCM = 2² × 3¹ × 5² = 4 × 3 × 25 = 300.
Thus, the smallest number divisible by 20, 25, and 30 is 300.
3. How do you solve questions on the divisibility test for 11 from NCERT Chapter 3? For example, is 5445 divisible by 11?
To check if a number is divisible by 11, you use the following method:
- Step 1: Find the sum of the digits at the odd places (from the right). For 5445, this is the first and third digit: 5 + 4 = 9.
- Step 2: Find the sum of the digits at the even places. For 5445, this is the second and fourth digit: 4 + 5 = 9.
- Step 3: Find the difference between these two sums. Here, the difference is 9 – 9 = 0.
- Step 4: If the difference is 0 or a multiple of 11, the number is divisible by 11. Since the result is 0, the number 5445 is divisible by 11.
4. Why is the prime factorisation method a reliable way to find both HCF and LCM?
The prime factorisation method is reliable because it breaks down numbers into their most basic, unique building blocks, which are the prime numbers.
- For the HCF, this method helps you clearly see the 'common' building blocks shared by all the numbers. By multiplying only these shared factors, you find the highest possible number that can divide all of them.
- For the LCM, it ensures you include every building block from all numbers at its highest required quantity (power). This guarantees the final result is the smallest number that contains all the original numbers as factors.
5. What is the correct method to solve word problems involving HCF and LCM in NCERT Class 6 Maths Chapter 3?
To solve word problems, you must first identify whether to calculate HCF or LCM by looking for keywords:
- Calculate HCF if the problem asks for the greatest, maximum, or largest number to divide or group different quantities into equal sets. For example, finding the 'maximum capacity' of a container that can measure the oil from different tankers an exact number of times.
- Calculate LCM if the problem asks for the smallest, minimum, or least quantity or the next time something will happen simultaneously. For example, finding when three bells, ringing at different intervals, will 'ring together' again.
Once identified, you can apply the respective method. You can find more solved examples in the NCERT Solution for Class 6 Maths Chapter 3.
6. What is the key difference between factors and multiples, and why are they commonly confused in Chapter 3?
The key difference lies in their relationship to a number:
- A factor is a number that divides another number exactly, leaving no remainder. Factors of a number are finite and are always less than or equal to the number itself. For example, the factors of 8 are 1, 2, 4, and 8.
- A multiple is the result of multiplying a number by an integer. Multiples of a number are infinite and are always greater than or equal to the number itself. For example, the multiples of 8 are 8, 16, 24, 32, and so on.
They are often confused because both involve multiplication and division. A helpful way to remember is: 'Factors are few, Multiples are many'.
7. How do you check if a large number is divisible by 3 using the method given in the NCERT solutions for Chapter 3?
The method to check for divisibility by 3 is simple and works for any number, no matter how large. Follow these steps:
- Step 1: Add up all the digits of the number. For instance, to check the number 15,963.
- Step 2: Sum the digits: 1 + 5 + 9 + 6 + 3 = 24.
- Step 3: Check if this sum is divisible by 3. Since 24 is divisible by 3 (24 ÷ 3 = 8), the original number is also divisible by 3.
- Conclusion: 15,963 is divisible by 3. If the sum was not divisible by 3, the original number would not be either.
8. Can two numbers have 15 as their HCF and 175 as their LCM? How do the solutions for Chapter 3 help us reason this?
No, two numbers cannot have 15 as their HCF and 175 as their LCM. The NCERT solutions for Chapter 3 are based on a fundamental property of numbers: the HCF of two numbers must always be a factor of their LCM. To check this:
- Step 1: Divide the proposed LCM by the proposed HCF.
- Step 2: Perform the division: 175 ÷ 15.
- Step 3: The result is 11 with a remainder of 10.
Since 15 does not divide 175 exactly, it is not a factor. Therefore, it is mathematically impossible for such a pair of numbers to exist.
9. How are prime and composite numbers defined in Chapter 3, and what is the correct way to identify them?
In Chapter 3, prime and composite numbers are defined based on their number of factors:
- Prime Numbers: These are natural numbers greater than 1 that have exactly two distinct factors: 1 and the number itself. Examples include 2, 3, 5, 7, 11.
- Composite Numbers: These are natural numbers greater than 1 that have more than two factors. Examples include 4 (factors 1, 2, 4), 6 (factors 1, 2, 3, 6), and 9 (factors 1, 3, 9).
To identify a number, you check how many factors it has. If it only has two, it's prime; if it has more than two, it's composite. For a deeper explanation, you can explore the concepts of Prime and Composite Numbers.
10. In the NCERT solutions for 'Number Play', why is the number 1 considered neither prime nor composite?
The number 1 is a special case and is considered neither prime nor composite because it doesn't fit the strict definition of either:
- The definition of a prime number requires it to have exactly two distinct factors (1 and itself). The number 1 has only one factor: 1.
- The definition of a composite number requires it to have more than two factors. The number 1 fails this condition as well.
Since 1 does not satisfy the criteria for being prime or composite, it is classified uniquely on its own.
