Get the simplified Class 6 Maths NCERT Solutions of Ganita Prakash Chapter 3 Number Play textbook exercise questions with complete explanation.

## Ganita Prakash Class 6 Maths Chapter 3 Solutions Number Play

**NCERT Solutions for Class 6 Maths Ganita Prakash Chapter 3 Number Play**

**3.1 Numbers can Tell us Things 3.2 Supercells Figure it Out (Page No. 57-58)**

Question 1.

Colour or mark the supercells in the table below.

Solution:

Question 2.

Fill the table below with only 4-digit numbers such that the supercells are exactly the c cells.

Solution:

Question 3.

Fill the table below such that we get as many supercells as possible. Use numbers between 100 and 1000 without repetitions.

Solution:

Question 4.

Out of the 9 numbers, how many supercells are there in the table above?

Solution:

Out of 9 numbers, there are 5 supercells in the above table.

Question 5.

Find out how many supercells are possible for different numbers of cells. Do you notice any pattern? What is the method to fill a given table to get the maximum number of supercells? Explore and share your strategy.

Solution:

If there are n odd cells then number of supercells = \(\frac{n+1}{2}\)

If there are n even cells then number of supercells = \(\frac{n}{2}\)

Yes, there is a pattern. Alternate cells can be supercells.

Method to fill a given table to get the maximum number of supercells.

- Make first cell as supercell. After that each alternate cell is to be made supercell.
- No consecutive cells can be supercell except in case of 4 cells because then first and fourth cell can be supercell.

Question 6.

Can you fill a supercell table without repeating numbers such that there are no supercells? Why or why not?

Solution:

No, it is not possible to fill a supercell table without repeating numbers such that there are no supercells.

As there are two cases:

Case I: If we fill the cells in descending order then the first cell be supercell.

Case II: If we fill the cells in ascending order then the last cell will be supercell.

If we don’t follow any order, then there will definitely atleast one supercell.

Question 7.

Will the cell having the largest number in a table always be a super cell? Can the cell having the smallest number in a table be a supercell? Why or why not?

Solution:

Yes, the cell having the largest number in a table always be a supercell because if it is comer cell, then the number adjacent to it (i.e. either second cell or second last cell) will be smaller than it. If it is in between then both its adjacent number would be smaller than it.

No, the cell having smallest number in a table can not be supercell because number adjacent to it will always be larger/greater than it.

Question 8.

Fill a table such that the cell having the second largest number is not a supercell.

Solution:

Here 980 is the second largest number but it is not a supercell as 999 is the supercell.

Question 9.

Fill a table such that the cell having the second largest number is not a supercell but the second smallest number is a super cell. Is it possible?

Solution:

Here 1870 is the second largest number but the cell having 1870 is not a supercell because number 1895 (adjacent to it) is greater than it.

489 is the second smallest number but the cell having 489 is a supercell because adjacent number 475 is smaller to it.

Question 10.

Make other variations of this puzzle and challenge your classmates.

Solution:

Fill a table such that only even numbers are supercell.

Fill a table such that all the supercells are divisible by 5.

Intext Question

Question 1.

Complete Table 2 with 5-digit numbers whose digits are ‘1’, ‘0’, ‘6’. ‘3’, and ‘9’ in some order. Only a coloured cell should have a number greater than all its neighbours. (Page 58)

Solution:

96,310 | 96,301 | 36,109 | 63,109 |

10,369 | 13,609 | 60,319 | 19,306 |

10,936 | 36,910 | 60,193 | 39,106 |

10,369 | 10,963 | 10,639 | 39,610 |

The biggest number in the table is 96, 310.

The smallest even number in the table is 10,936.

The smallest number greater than 50,000 in the table is 60,193.

**3.3 Patterns of Numbers on the Number Line Figure it Out (Page No. 59)**

Question 1.

Identify the numbers marked on the number lines below, and label the remaining positions.

Put a circle around the smallest number and a box around the largest number in each of the sequences above.

Answer:

**3.4 Playing with Digits (Page No. 60)**

Question 1.

Digit sum 14

(a) Write other numbers whose digits add up to 14.

(b) What is the smallest number whose digit sum is 14?

(c) What is the largest 5-digit whose digit sum is 14?

(b) How big a number can you form having the digit sum 14? Can you make an even bigger number?

Solution:

(a) Some numbers whose digits add up to 14 are:

59, 68, 77, 86, 95, 149, 158, 167, 176, 185, 194, 239, 248, 257, 266, 275, 281, 293

(b) The smallest number whose digit sum is 14 = 59.

(c) The largest 5 digit number containing 0 whose digit sum is 14 = 95,000.

The largest 5 digit number not containing 0 whose digit sum is 14 = 92,111.

(d) A very big number having the digit sum 14 can be made. e.g. 95000000000000.

Yes, we can make even bigger number e.g. 9500000000000000000000.

Question 2.

Find out the digit sums of all the numbers from 40 to 70. Share your observations with the class.

Solution:

Number | Digit Sum |

40 | 4 |

41 | 5 |

42 | 6 |

43 | 7 |

44 | 8 |

45 | 9 |

46 | 10 |

47 | 11 |

48 | 12 |

49 | 13 |

50 | 5 |

51 | 6 |

52 | 7 |

53 | 8 |

54 | 9 |

55 | 10 |

56 | 11 |

57 | 12 |

58 | 13 |

59 | 14 |

60 | 06 |

61 | 1 |

62 | 8 |

63 | 9 |

64 | 10 |

65 | 11 |

66 | 12 |

67 | 13 |

68 | 14 |

69 | 15 |

70 | 7 |

Question 3.

Calculate the digit sums of 3-digit numbers whose digits are consecutive (for example, 345). Do you see a pattern? Will this pattern continue?

Solution:

If we take numbers in reverse order, sum of digits will remain same.

Yes, we observe a pattern.

i. e. (first number + 1) × 3 = digit sum.

Intext Question

Question 1.

Among the numbers 1-100, how many times will the digit ‘7’occur?

Solution:

The total count of 7 that we get is 20.

Question 2.

Among the numbers 1-1000, how many times will the digit ‘7’occur?

Solution:

The number of times 7 will be written when listing the numbers from 1 to 1000 is 300.

**3.5 Pretty Palindromic Patterns 3.6 The Magic Number of Kaprekar 3.7 Clock and Calendar Numbers Figure it Out (Page No. 64-65)**

Question 1.

Pratibha uses the digits ‘4‘7’, ‘3’and ‘2’ and makes the smallest and largest 4-digit numbers with them: 2347 and 7432. The difference between these two numbers is 7432 – 2347 = 5085. The sum of these two numbers is 9779. Choose 4-digits to make:

(a) the difference between the largest and smallest numbers greater than 508$.

(b) the difference between the largest and smallest numbers less than 5085.

(e) the sum of the largest and smallest numbers greater than 9779.

(d) the sum of the largest and smallest numbers less than 9779.

Solution:

(a) Digits – 8, 7, 3 and 2

Largest Number – 8 7 3 2

Smallest Number – 2 3 7 8

Difference = 6 3 5 4

6354 > 5085

(b) Digits – 1, 2, 3 and 4

Largest Number – 4 3 2 1

Smallest Number – 1 2 3 4

Difference = 3 0 8 7

3087 < 5085

(c) Digits – 9, 8, 7 and 6

Largest Number – 9 8 7 6

Smallest Number – 6 7 8 9

Sum = 166 6 5

16665 > 9779

(d) Digits – 1, 2, 3 and 8

Largest Number 8 3 2 1

Smallest Number 1 2 3 8

Sum = 9 5 5 9

9559 < 9779

Question 2.

What is the sum of the smallest and largest 5-digit palindrome? What is their difference?

Solution:

Case-I. Smallest 5 digit palindrome number (different digits) – 1 2 3 2 1

Largest 5 digit palindrome number (different digits) 98789

Sum = 12321 + 98789 = 111110

Difference = 98789 – 12321

= 86468

Case-II: Largest 5 digit palindrome (same digit) – 99999

Smallest 5 digit palindrome (same digit) – 11111

Sum – 99999 + 11111 = 111110

Difference = 99999 – 11111

= 88888

Question 3.

The time now is 10:01. How many minutes until the clock shows the next palindromic time? What about the one after that?

Solution:

Time now – 10 : 01

Now next palindrome time is 11 : 11

Hence, 11:11 – 10:01 = 70 minutes.

Question 4.

How many rounds does the number 5683 take to reach the Kaprekar constant?

Intext Questions

Question 1.

Write all possible 3-digit palindromes using these digits 1,2,3. (Page 61)

Solution:

- 111
- 121
- 131
- 212
- 222
- 232
- 313
- 323
- 333

Question 2.

Will reversing and adding numbers repeatedly, starting with a 2-digit number, always give a palindrome? Explore and find out. (Page 62)

Solution:

All two-digit numbers eventually become palindromes after repeated reversal and addition. About 80% of all numbers under 10,000 resolves into a palindrome ip four or fewer steps; about 90% of those resolve in seven steps or fewer.

Example 1: Number 12

1. Initial Number: 12

2. Reverse: 21

3. Add: 12 + 21 = 33

4. Palindromp

Check: 33 is a palindrome.

• Result: 33 is a palindrome.

Example 2: Number 89

1. Initial Number: 89

2. Reverse: 98

3. Add: 89 + 98 = 187

4. Palindrome Check: 187 is not a palindrome.

5. Reverse 187: 781

6. Add: 187 + 781 =968

7. Palindrome Check: 968 is not a palindrome.

Puzzle time (Page 62)

I am a 5-digit palindrome.

I am an odd number.

My ‘t’ digit is double of my ‘u’ digit.

My ‘h’ digit is double of my ‘t’ digit. Who am I?

Solution:

12421

Explore (Page 63)

Take different 4-digit numbers and try carrying out these steps. Find out what happens.

Solution:

Selected a 4-digit number 1234

- Starting number: 1234
- Descending order: 4321
- Ascending order: 1234
- Subtract: 4321 – 1234 = 3087

Repeat:

- Descending order of 3087: 8730
- Ascending order of 3087: 0378
- Subtract: 8730 – 0378 = 8352

Repeat:

- Descending order of 8352: 8532
- Ascending order of 8352: 2358
- Subtract: 8532 – 2358 = 6174

Result: 6174 (Kaprekar constant)

Question 1.

Carry out these same steps with a few 3-digit numbers. What number will start repeating? (Page 63)

Solution:

We will do this with help of two examples:

1. Number 123

- Starting number: 123
- Descending order: 321
- Ascending order: 123
- Subtract: 321 – 123 = 198

Repeat:

- Descending order of 198: 981
- Ascending order of 198: 189
- Subtract: 981 – 189 = 792

Repeat:

- Descending order of 792: 972
- Ascending order of 792: 279
- Subtract: 972 – 279 = 693

Repeat:

- Descending order of 693: 963
- Ascending order of 693: 369
- Subtract: 963 – 369 = 594′

Repeat:

- Descending order of 594: 954
- Ascending order of 594: 459
- Subtract: 954 – 459 = 495

Repeat:

- Descending order of 495: 954
- Ascending order of 495: 459
- Subtract: 954 – 459 = 495

Result: The number 495 starts repeating.

2. Now let’s take the number 317

- Starting number: 317
- Descending order: 731
- Ascending order: 137
- Subtract: 731 – 137 = 594

Repeat:

- Descending order of 594: 954
- Ascending order of 594: 459
- Subtract: 954 – 459 = 495 ,

Repeat:

- Descending order of 495: 954
- Ascending order of 495: 459
- Subtract: 954 – 459 = 495

Result: The number 495 starts repeating. When applying Kaprekar’s routine to 3- digit numbers, the number 495 is often reached and starts repeating. This number is known as the Kaprekar constant for 3-digit numbers.

Try and find out all possible times on a 12-hour clock of each of these types. For example, 4:44, 10:10, 12:21. (Page 64)

Solution:

01:10, 02:20, 03:30, 04:40, 05:50, 10:01, 11:11, 12:21

Find some other dates of this form from the past like 20/12/2012 where the digits ‘2’, ‘0’, ‘1 ’, and ‘2 ’ repeat in that order. (Page 64)

Solution:

11/02/2011, 22/02/2022, 01/10/2010, 10/01/2010, 02/02/2020

**3.8 Mental Math Figure it Out (Page No. 66 – 67)**

Question 1.

Write an example for each of the below scenarios whenever possible.

Could you find examples for all the cases? If not, think and discuss what could be the reason. Make other such questions and challenge your classmates.

(a) Let’s divide

90, 250 by 2

then \(\frac{90,250}{2}\) = 45,125

Hence to get sum more than 90,250 both numbers should be more than 45,125.

(b) To get a 6 digit sum by adding 5 digit and 3 digit, the 5 digit number should be more than 99,001.

(c) Let’s take minimum 4 digit number 1000

let’s add them

which is a 4 digit number.

Hence 6 digit sum’ from 4 digit number is impossible.

(d) Let’s take 5 digit numbers 67987 and 65783

let’s add them

which is a 6 digit number.

(e) Let’s take minimum 5 digit numbers 1000

let’s add them

which is a 5 digit number.

Hence 6 digit sum from 5 digit numbers is impossible.

(f) 5-digit -5

digit to

give a difference less than 56,503

< 56503

(g) 5-digit-3 digit = 1 008 6 (5 digit)

to give a 4 digit = + 875 (3 digit)

difference = 92 11 (4 digit)

(h) 5-digit digit = 1 2 8 7 6 (5 digit)

to give a 4 digit = -7865 (4 digit)

difference = 5 0 11 (4 digit)

(i) 5-digit -5 digit 7 = 645 3 (5 digit)

to give a 3 digit = 76 145 (5 digit)

difference = 308 (3 digit)

(j) 5-digit -5 digit Not possible to give 91,500

Question 2.

Always, Sometimes, Never?

Below are some statements. Think, explore and find out if each of the statement is ‘Always true’, ‘Only sometimes true’ or ‘Never true’. Why do you think so? Write your reasoning; discuss this with the class.

(a) 5-digit number + 5-digit number gives a 5-digit number

(b) 4-digit number + 2-digit number gives a 4-digit number

(c) 4-digit number + 2-digit number gives a 6-digit number

(d) 5-digit number – 5-digit number gives a 5-digit number

(e) 5-digit number – 2-digit number gives a 3-digit number

Solution:

**3.9 Playing with Number Patterns 3.10 An Unsolved Mystery – the Collatz Conjecture! 3.11 Simple Estimation 3.12 Games and Winning Strategies Figure it Out (Page No. 72 – 73)**

Question 1.

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

Solution:

If we swap first and last digit of central number 62,871, we get 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 reacall 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.

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

1. 64 is even, divide by 2 =32

2. 32 is even, divide by 2 = 16

3. 16 is even, divide by 2 = 8

4. 8 is even, divide by 2 = 4

5. 4 is even, divide by 2 = 2

6. 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.

Intext Questions

Example: In the following, there is a number pattern on +3 being followed.

Find out the sum of the numbers in each of the below figures. Should we add them one by one or can we use a quicker way? Share and discuss in class the different methods each of you used to solve these questions. (See figures, NCERT TB, Pages 67-68)

Solution:

(a) In figure (a), number 40 is repeated 12 times and number 50 is repeated 10 times

Hence sum of all numbers = 40 × 12 + 50 × 10

= 480 + 500 = 980

(b) In figure (b), 1 dot (•) is 44 times and 5 dots (•) are 20 times

Hence sum of all dots = 1 × 44 + 5 × 20 = 44 + 100 = 144

(c) In figure (c), number 32 is 32 times and number 64 is 16 times

Hence sum of all numbers = 32 × 32 + 64 × 16 = 1024 + 1024 = 2048

(d) In figure (d), 3 dots (•) are 17 times and 4 dots (•) are 18 times

Hence sum of all dots = 17 × 3 + 18 × 4 = 51 + 72 = 123

(e) In figure (e), number 15 is 22 times, number 25 is 22 times and number 35 is 22 times

Hence sum of all numbers = 15 × 22 + 25 × 22 + 35 × 22 = 330 + 550 + 770 = 1650

(f) In figure (f), number 125 is 18 times, number 250 is 8 times and number 500 is 4 times and number 1000 is one time.

Hence sum of all numbers = 125 × 18 + 250 × 8 + 500 × 4 + 1000 = 2250 + 2000 + 2000 +1000 = 7250