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NCERT Solutions For Class 6 Maths Chapter 3 Number Play Exercise 3.12 - 2025-26

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Class 6 Maths Chapter 3 Questions and Answers - Free PDF Download

In NCERT Solutions for Class 6 Maths Chapter 3 Number Play Exercise 3.12, you’ll dive into fun games and clever tricks that make maths interesting. This chapter teaches you how to spot patterns, use logic, and find winning strategies, all while playing with numbers and simple puzzles! If you feel confused about tricky number patterns or want to get better at predicting game outcomes, you’ll find helpful step-by-step answers here.

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Vedantu’s NCERT Solutions are easy to understand and specially made to help you prepare for your exams with confidence. You can also download free PDFs to practise anytime and improve your problem-solving skills. Explore the full CBSE Class 6 Maths Syllabus for more details.


Use these NCERT Solutions for Class 6 Maths to build a strong base in mathematical thinking and get ahead in your revision. For more practice and extra study material, check out the Class 6 Maths NCERT Solutions collection.


Access NCERT Solutions for Class 6 Maths Ex 3.12 Number Play

3.12 Games and Winning Strategies 

Figure it Out (Page No. 72 – 73)

Question 1. This grid has only one supercell (number greater than all its neighbours). If you exchange two digits of one of the numbers, there will be 4 supercells. Figure out which digits to swap.


16,200

39,344

29,765

23,609

62,871

45,306

19,381

50,319

38,408



Solution:
If we swap first and last digit of central number 62,871, we get the desired result.


16,200

39,344

29,765

23,609

21,876

45,306

19,381

50,319

38,408



Question 2. How many rounds does your year of birth take to recall the Kaprekar constant?
Solution: If your year of birth is 2000
Step 1: Now from digits of number 2000
Here largest number = 2000
and smallest number = 0002
Let’s subtract them = 2000 – 0002 = 1998


Step 2: Now from digits of number 1998
Here largest number = 9981
and smallest number = 1899
Let’s subtract them = 9981 – 1899 = 8082


Step 3: Now from digits of number 8082
Here largest number = 8820
and smallest number = 0288
Let’s subtract them = 8820 – 0288 = 8532


Step 4: Now from digits of number 8532
Here largest number = 8532
and smallest number = 2358

Let’s subtract them = 8532 – 2358 = 6174


which is a Kaprekar constant.
Hence it took 4 rounds to reach the Kaprekar constant from 2000.


Question 3. We are the group of 5-digit numbers between 35,000 and 75,000 such that all of our digits are odd. Who is the largest number in our group? Who is the smallest number in our group? Who among us is the closest to 50,000?
Solution

The largest number with all odd digits (different) = 73951
The largest number with all odd digits (repetitive) = 73999
The smallest number (non repetitive) = 35,179
The smallest number (repetitive) = 57111
Closest to 50,000 (in case of non-repetition) = 49751
Closest to 50,000 (in case of repetition) = 49999


Question 4. Estimate the number of holidays you get in a year including weekends, festivals and vacation. Then try to get an exact number and see how close your estimate is.
Solution:
Will be done by students.


Question 5. Estimate the number of litres a mug, a bucket and an overhead tank can hold.
Solution:
Will be done by students


Question 6. Write one 5-digit number and two 3-digit numbers such that their sum is 18,670. ‘
Solution:
5 digit number = 1 8 0 0 0
3 digit number = 6 7 0
Sum = 1 8 000 + 670 = 18670


Question 7. Choose a number between 210 and 390. Create a number pattern similar to those shown in Section 3.9 that will sum up to this number.


Choose a number between 210 and 390. Create a number pattern similar to those shown in Section 3.9 that will sum up to this number


Solution:
Sum of No. = 5 × 1 = 5
+ 10 × 3 = 30
+ 15 × 5 = 75
+ 20 × 7 = 140 = 250
which lies between 210 and 390.


Question 8. Recall the sequence of Powers of 2 from Chapter 1, Table 1. Why is the Collatz conjecture correct for all the starting numbers in this sequence?
Solution: The square of power of 2 is :
1,2,4, 8, 16, 32, 64

Let’s take the number’ 64 as per Collatz Conjecture


  • 64 is even, divide by 2 = 32

  • 32 is even, divide by 2 = 16

  • 16 is even, divide by 2 = 8

  • 8 is even, divide by 2 = 4

  • 4 is even, divide by 2 = 2

  • 2 is even, divide by 2 = 1


Hence Collatz conjecture is correct in all numbers in the power of 2 sequence.
As it is power of 2, and in Collatz Conjecture even number is divided by 2 in each step.


Question 9. Check if the Collatz Conjecture holds for the starting number 100.
Solution:

  • 100 is even, divide by 2 =50

  • 50 is even, divide by 2 = 25

  • 25 is odd, so multiply by 3 and add 1 → 76

  • 76 is even, divide by 2 = 38

  • 38 is even, divide by 2 = 19

  • 19 is odd, so multiply by 3 and add 1 → 58

  • 58 is even, divide by 2 = 29

  • 29 is odd, so multiply by 3 and add 1 → 88

  • 88 is even, divide by 2 = 44

  • 44 is even, divide by 2 = 22

  • 22 is even, divide by 2 = 11

  • 11 is odd, so multiply by 3 and add 1 → 34

  • 34 is even, divide by 2 = 17 

  • 17 is odd, so multiply by 3 and add 1 → 52

  • 52 is even, divide by 2 = 26

  • 26 is even, divide by 2 = 13

  • 13 is odd, so multiply by 3 and add 1 → 40

  • 40 is even, divide by 2 = 20

  • 20 is even, divide by 2 = 10

  • 10 is even, divide by 2 = 5

  • 5 is odd, so multiply by 3 and add 1 → 16

  • 16 is even, divide by 2 = 8

  • 8 is even, divide by 2 = 4

  • 4 is even, divide by 2 = 2

  • 2 is even, divide by 2 = 1


Yes, the Collatz conjecture holds for the starting number 100. 


Benefits of NCERT Solutions for Class 6 Maths Ex 3.12

  • Enhances problem-solving skills: By learning to analyze game strategies, students improve their ability to solve problems logically.

  • Improves reasoning ability: The chapter sharpens the reasoning skills of students by teaching them to recognize patterns and predict outcomes.

  • Practical application of maths: Students learn how maths can be applied to real-life situations through games.

  • Boosts critical thinking: The chapter encourages critical thinking as students explore various strategies to win games.

 

Class 6 Maths Chapter 3: Exercises Breakdown

Exercise

Topic

Exercise 3.1

Numbers Can Tell Us Things

Exercise 3.2

Supercells

Exercise 3.3

Patterns of Numbers on the Number Line

Exercise 3.4

Playing with Digits

Exercise 3.5

Pretty Palindromic Patterns

Exercise 3.6

The Magic Number of Kaprekar

Exercise 3.7

Clock and Calendar Numbers

Exercise 3.8

Mental Math

Exercise 3.9

Playing with Number Patterns

Exercise 3.10

An Unsolved Mystery — the Collatz Conjecture!

Exercise 3.11

Simple Estimation



Important Study Material Links for Maths Chapter 3 Class 6

S.No.

Important Study Material Links for Chapter 3

1.

Class 6 Number Play Important Questions

2.

Class 6 Number Play Notes

3.

Class 6 Maths Number Play Worksheets



Conclusion

Chapter 3 Excercise 3.12  Games and Winning Strategies from Class 6 Maths is an engaging and interactive way to strengthen mathematical concepts. Students not only enjoy the process of learning but also understand how to apply mathematical reasoning in real-life scenarios. This chapter is instrumental in building logical thinking and problem-solving abilities through fun and strategic games.


Chapter-wise NCERT Solutions Class 6 Maths

After familiarising yourself with the Class 6 Maths  Chapters Question Answers, you can access comprehensive NCERT Solutions from all Class 6 Maths textbook chapters.




Related Important Links for Class 6  Maths 

Along with this, students can also download additional study materials provided by Vedantu for Maths  Class 6-


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FAQs on NCERT Solutions For Class 6 Maths Chapter 3 Number Play Exercise 3.12 - 2025-26

1. Where can I find reliable and step-by-step NCERT Solutions for Class 6 Maths Chapter 3, Number Play, for the 2025-26 session?

You can find comprehensive and accurate NCERT Solutions for Class 6 Maths Chapter 3 Number Play on Vedantu. These solutions are prepared by subject matter experts and follow the latest CBSE 2025-26 guidelines, providing clear, step-by-step explanations for every question in the textbook exercises to help you build a strong conceptual foundation.

2. What is the correct method to find the HCF of two numbers using prime factorisation as shown in Chapter 3?

To find the Highest Common Factor (HCF) using the prime factorisation method, you should follow these steps:

  • Step 1: Find the prime factors of each number separately.
  • Step 2: Identify the common prime factors from the lists of factors for all the numbers.
  • Step 3: Multiply these common prime factors together. The product is the HCF of the given numbers. For example, for 12 and 18, the prime factors are 2×2×3 and 2×3×3. The common factors are 2 and 3, so the HCF is 2×3 = 6.

3. How do you solve for the LCM of numbers using the common division method in Class 6 Maths Chapter 3?

The common division method is an efficient way to find the Lowest Common Multiple (LCM). Here is the step-by-step process:

  • Step 1: Arrange the given numbers in a row, separated by commas.
  • Step 2: Divide the numbers by the smallest prime number that can divide at least one of them.
  • Step 3: Write the quotients and any undivided numbers in the row below.
  • Step 4: Repeat this process until all the numbers in the last row become 1.
  • Step 5: The product of all the prime divisors you used is the LCM of the numbers.

4. How do the NCERT Solutions on Vedantu make solving problems from Chapter 3, Number Play, easier?

Vedantu's NCERT Solutions for Chapter 3 simplify learning by breaking down complex problems into easy-to-understand steps. They explain the logic behind divisibility tests, finding HCF and LCM, and handling word problems. Each solution is crafted to align with the CBSE marking scheme, ensuring you learn the correct method to score full marks in exams. For a complete understanding, you can refer to the Class 6 Maths Chapter 3 Number Play Notes which summarises all key concepts.

5. What is the fundamental difference between HCF and LCM, and how do I know which one to use for word problems in this chapter?

The key difference is that HCF (Highest Common Factor) is the largest number that can divide a set of numbers without leaving a remainder, while LCM (Lowest Common Multiple) is the smallest number that is a multiple of all numbers in the set. Use HCF when a problem asks for finding the largest item to 'split' or 'distribute' things into equal groups. Use LCM when a problem involves finding a common time or point when events will happen 'together again'.

6. Why is the number 1 considered neither a prime number nor a composite number in mathematics?

The number 1 is unique because it has only one factor: itself. The definition of a prime number requires it to have exactly two distinct factors (1 and itself). The definition of a composite number requires it to have more than two factors. Since 1 does not fit either of these definitions, it is classified as neither prime nor composite. You can explore more about prime and composite numbers to understand their properties better.

7. How do I solve the problems in Exercise 3.7 of Chapter 3, which often involve real-life applications of HCF and LCM?

The word problems in Exercise 3.7 require you to first identify whether to calculate HCF or LCM. For questions asking about maximum capacity, longest tape, or arranging items in groups, you typically need to find the HCF. For questions about finding when things will happen simultaneously again (like bells ringing together), you need to find the LCM. For detailed, stepwise answers, you can refer to Vedantu's NCERT Solutions for Class 6 Maths Ch 3 Number Play Ex 3.7.

8. What is the correct way to apply the divisibility test for 11 as per the NCERT textbook?

To check if a number is divisible by 11, follow this method:

  • Step 1: Find the sum of the digits at the odd places (from the right).
  • Step 2: Find the sum of the digits at the even places (from the right).
  • Step 3: Calculate the difference between these two sums.
  • Step 4: If the difference is either 0 or a multiple of 11, the number is divisible by 11. Otherwise, it is not.