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 coloured cells.
Solution:
Question 3.
Fill in the table below such that we get as many supercells as possible. Use numbers between 100 and 1000 without repetitions.
Solution:
Maximum number of supercells that we can get is 5.
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:
Question 6.
Can you fill a supercell table without repeating numbers such that there are no supercells? Why or why not?
Solution:
No, if we fill with numbers increasing by 1, we get at least one supercell, i.e., the last number.
Question 7.
Will the cell having the largest number in a table always be a supercell? Can the cell having the smallest number in a table be a supercell? Why or why not?
Solution:
Yes, the cell with the largest number in a table will always be a supercell. Since the largest number is greater than any other number in the table, it will naturally be greater than its adjacent numbers. Therefore, it will always be a supercell.
No, the cell with the smallest number in a table cannot be a supercell. Since the smallest number in the table is less than or equal to all other numbers, it cannot be greater than its adjacent numbers. Thus, it cannot be a supercell.
Question 8.
Fill a table such that the cell having the second largest number is not a supercell.
Solution:
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 supercell. Is it possible?
Solution:
Yes, this is possible.
Question 10.
Make other variations of this puzzle and challenge your classmates.
Solution:
Fill a table such that the cell having the second smallest number is not a supercell.
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.
The biggest number in the table is _____________.
The smallest even number in the table is _____________.
The smallest number greater than 50,000 in the table is _____________
Solution:
Once you have filled the table above, put commas appropriately after the thousands digit.
| 96,310 | 96,301 | 36,109 | 36,910 |
| 13,690 | 13,609 | 60,319 | 19,306 |
| 13,906 | 10,936 | 60,193 | 19,360 |
| 10,369 | 10,963 | 10,396 | 19,630 |
The biggest number in the table is 96,310.
The smallest even number in the table is 10,396.
The smallest number greater than 50,000 in the table is 60,193. (Answer may vary)
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?
(d) How big a number can you form having the digit sum 14? Can you make an even biggermumber?
Solution:
(a) 590, 770, 248, 680 etc.
(b) 59, (∵ 5 + 9 = 14)
(c) 95000,(∵ 9 + 5 + 0 + 0 + 0 = 14)
(d) Infinite numbers can be formed having the digit sum 14.
Yes, we can make an bigger even number.
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 + 0 = 4 |
| 41 | 4 + 1 = 5 |
| 42 | 4 + 2 = 6 |
| 43 | 4 + 3 = 7 |
| 44 | 4 + 4 = 8 |
| 45 | 4 + 5 = 9 |
| 46 | 4 + 6 = 10 |
| 47 | 4 + 7 = 11 |
| 48 | 4 + 8 = 12 |
| 49 | 4 + 9 = 13 |
| 50 | 5 + 0 = 5 |
| 51 | 5 + 1 = 6 |
| 52 | 5 + 2 = 7 |
| 53 | 5 + 3 = 8 |
| 54 | 5 + 4 = 9 |
| 55 | 5 + 5 = 10 |
| 56 | 5 + 6 = 11 |
| 57 | 5 + 7 = 12 |
| 58 | 5 + 8 = 13 |
| 59 | 5 + 9 = 14 |
| 60 | 6 + 0 = 6 |
| 61 | 6 + 1 = 7 |
| 62 | 6 + 2 = 8 |
| 63 | 6 + 3 = 9 |
| 64 | 6 + 4 = 10 |
| 65 | 6 + 5 = 11 |
| 66 | 6 + 6 = 12 |
| 67 | 6 + 7 = 13 |
| 68 | 6 + 8 = 14 |
| 69 | 6 + 9 = 15 |
| 70 | 7 + 0 = 7 |
Observation:
From 40 to 49: The digit sum increased from 4 to 13
From 50 to 59: The digit sum increased from 5 to 14
From 60 to 69: The digit sum increased from 6 to 15
From 70 to 79: The digit sum will increase from 7 to 16
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:
1 + 2 + 3 = 6,
2 + 3 + 4 = 9,
3 + 4 + 5= 12,
4 + 5 + 6 = 15,
5 + 6 + 7 = 18,
6 + 7 + 8 = 21,
7 + 8 + 9 = 24.
Yes, the pattern is a table of 3 from 3 × 2 to 3 × 8.
We cannot continue this pattern after 789.
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 5085.
(b) the difference between the largest and smallest numbers less than 5085.
(c) 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) 5, 8, 3, 2; largest 4-digit number: 8532, smallest 4- digit number: 2358 Difference: 8532 – 2358 = 6174 > 5085
(b) 4, 6, 3, 2; largest 4-digit number: 6432, smallest 4- digit number = 2346 Difference: 6432 – 2346 = 4086 < 5085
(c) 5, 8, 3, 2 → 8532 + 2358 = 10890 > 9779
(d) 4, 6, 3, 2 → 6432 + 2346 = 8778 < 9779
(Answers may vary)
Question 2.
What is the sum of the smallest and largest 5-digit palindrome? What is their difference?
Solution:
Smallest 5-digit palindrome = 10001
Largest 5-digit palindrome = 99999
Sum of the smallest and largest 5-digit palindrome is 10001 + 99999 = 110000
The difference between the largest and smallest 5-digit palindrome is 99999 – 10001 = 89998
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:
Prime time = 10:01
Next palindromic time after 10:01 = 11:11
Difference = 11:11-10:01 = 70 minutes
Thus, the next palindromic time shows after 70 minutes.
Next palindromic time after 11 : 11=12:21
Difference = 12:21 – 11:11 = 70 minutes
= 1 hour 10 minutes
Thus, the next palindromic time shows after 1 hour 10 minutes and next one is 12 : 21 which comes total 2 hours 20 minutes later.
Question 4.
How many rounds does the number 5683 take to reach the Kaprekar constant?
Solution:
1st round,
A = 8653
B = 3568
C =8653 – 3568 = 5085
2nd round,
A = 8550
B = 0558
C = 8550 – 0558 = 7992
3rd round,
A = 9972
B = 2799
C = 9972 – 2799 = 7173
4th round,
A = 7731
B = 1377
C= 7731 – 1377 = 6354
5th round,
A = 6543
B = 3456
C= 6543 – 3456 = 3087
6th round,
A = 8730
B = 0378
C = 8730 – 0378 = 8352
7th round,
A = 8532
B = 2358
C = 8532 – 2358 = 6174
Thus, the number 5683 reaches the Kaprekar’s constant 6174 after 7 rounds.
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.
Solution:
| 5-digit + 5-digit to give a 5-digit sum more than 90,250 | 1st 5-digit number: 23,456 2nd 5-digit number: 66,795 Sum = 23,456 + 66,795 = 90251 >90250. |
| 5-digit + 3-digit to give a 6-digit sum | 5-digit number: 99,456 3-digit number: 795 Sum = 99,456 + 795 = 1,00,251 (a 6-digit sum). |
| 4-digit + 4-digit to give a 6-digit sum | Not possible, because if we take both the numbers as largest 4-digit number 9999, then the sum = 9999 + 9999 = 19,998; a 5-digit sum. |
| 5-digit + 5-digit to give a 6-digit sum | 1st 5-digit number: 63,456 2nd 5-digit number: 66,795 Sum = 63,456 + 66,795 = 1,30,251; a 6-digit sum. |
| 5-digit + 5-digit to give 18,500 | Not possible, because if we take the 1st 5-digit number as the smallest 5-digit number, i.e., 10,000, then, the second number will be 8,500 to get the required sum; which is a 4-digit number. |
| 5-digit – 5-digit to give a difference less than 56,503 | 1st 5-digit number: 93,456 2nd 5-digit number: 36,995 Difference = 56,461 < 56,603. |
| 5-digit – 3-digit to give a 4-digit difference | 5-digit number: 10,000 3-digit number: 999 Difference = 10,000 – 999 = 9,001; a 4-digit difference. |
| 5-digit – 4-digit to give a 4-digit difference | 5-digit number: 10,000 4-digit number: 1,000 Difference = 10,000 – 1,000 = 9,000; a 4-digit difference. |
| 5-digit – 5-digit to give a 3-digit difference | 1st 5-digit number: 60,456 2nd 5-digit number: 60,195 Difference = 60,456 – 60,195 261, a 3-digit difference. |
| 5-digit – 5-digit to give 91,500 | Not possible, because if we get the required difference we take the 1st 5-digit number, that will be the largest 5-digit number, i.e., 99,999 and the second number will be a 4-digit number ‘8499’. |
(Answer may vary)
For further do it yourself.
Question 2.
Always, Sometimes, Never?
Below are some statements. Think, explore and find out if each of the statements is ‘Always true’, ‘Only sometimes true’ or ‘Never true’. Why do you think so? Write your reasoning and 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:
(a) Only sometimes true.
The given statement is ‘5-digit number + 5-digit number gives a 5-digit number’. It is only sometimes true.
e.g. 10000 +10000 = 20000 i.e. 5-digit number
and 99999 + 99999 = 199998 i.e. 6-digit number
(b) The given statement is ‘4-digit number + 2-digit number gives a 4-digit number’. It is only sometimes true.
e.g. 1000 +10 = 1010 i.e. 4-digit number
and 9999 +10 = 10009 i.e. 5-digit number
(c) The given statement is ‘4-digit number + 2-digit number gives a 6-digit number’. It is never true.
e.g. 9999 + 99 = 10098 i.e. 5-digit number
(d) The given statement is ‘5-digit number – 5-digit number gives a 5-digit number’.
It is only sometimes true.
e.g. 99999 -10000 = 89999 i.e. 5-digit number
and 98765 – 94321 = 4444 i.e. 4-digit number
(e) The given statement is ‘5-digit number – 2-digit number gives a 3-digit number’. It is never true,
e.g. 10000 – 99 = 9901 i.e. 4-digit number.
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 the digits, 6 and 1 of 62,871, we get the 4 supercells.
Question 2.
How many rounds does your year of birth take to reach the Kaprekar constant?
Solution:
Do it yourself.
Question 3.
We are the group of 5-digit numbers between 35000 and 75000 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 50000?
Solution:
The odd numbers between 35000 and 75000 are
35001, 35003, 35005, ……..74999.
Therefore, largest number = 74999,
smallest number = 35001
and closest to 50000 = 49999 or 50001
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:
Number of holidays = 170 [estimate]
Number of holidays = 166 [exact]
Question 5.
Estimate the number of litres a mug, a bucket and an overhead tank can hold.
Solution:
Mug = 0.35 litres
Bucket = 20 litres
Overhead tank = 2000 litres.
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
- 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:
As per Collatz Conjecture rule: starts with any number; if the number is even, take half of it; if the number is odd, then multiply it by 3 and add 1; and repeat.
The sequence formed with starting number 100 is as follows:
100, 50, 25, 76, 38, 19, 58, 29, 88, 44, 22, 11, 34, 17, 52, 26, 13, 40, 20, 10, 5, 16, 8, 4, 2, 1.
Hence, 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