NCERT Solutions for Class 6 Maths Chapter 5 Prime Time

The name of Chapter 5 of Class 6 Maths in the Ganita Prakash curriculum is “Prime Time.” This chapter helps students understand the parts of numbers, such as factors and multiples. In Maths Class 6 Chapter 5, students learn about prime and composite numbers, divisibility rules, and methods to find common factors and multiples. These concepts are very important for solving higher-level maths problems and understanding number patterns.

To understand Chapter 5 Class 6 Maths Ganita Prakash, students should already know basic multiplication and division of whole numbers. The chapter title Prime Time goes beyond simple counting and focuses on what makes numbers unique. Students clearly learn the difference between prime numbers, which have only two factors (1 and the number itself), and composite numbers, which have more than two factors.

Chapter 5 of NCERT Class 6 Maths solutions focuses on applying concepts to real-life situations. Students learn how to check whether a number is divisible by 2, 3, 5, or 10 without doing long division. These divisibility tests save time and are very useful in later topics like HCF (Highest Common Factor), LCM (Lowest Common Multiple), and fraction simplification.

The Chapter 5 Class 6 Maths NCERT solutions explain every step clearly so students can check and understand their answers easily. Whether it is finding prime factorisation using factor trees or listing multiples within a given range, the solutions help students build strong logical thinking. For exam preparation and practice, these NCERT chapter 5 class 6 prime time question answer resources are very helpful, as they provide step-by-step explanations and improve confidence in solving maths problems.

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Chapter 5 Class 6 Maths Prime Time Question Answers

5.1 Common Multiples and Common Factors Figure it Out (Page No. 108)

Question 1.
At what number is ‘idli-vada’ said for the 10th time?
Solution:
The first number for which the players should say, ‘idli-vada’ is 15 as 3 × 5, which is a multiple of 3 and 5.
So, 15 × 1 = 15,
15 × 2 = 30,
15 × 3 = 45,
………………….
………………….
………………….

15 × 10 = 150

Thus, 150 is the number that will be said ‘idli-vada’ for the 10th time.

Question 2.
If the game is played for the numbers from 1 till 90, find out:
(i) How many times would the children say ‘idli’ (including the times they say ‘idli-vada’)?
(ii) How many times would the children say ‘vada’ (including the times they coy ‘idli-vada’)?
(iii) How many times would the children say ‘idli-vada’?

Solution:
(i) The multiples of 3 from 1 till 90 are 3, 6, 9, 12,15,18, 21, 24, 27, 30, 33, 36, 39,42,45,48, 51, 54, 57, 60, 63, 66, 69, 72, 75, 78, 81, 84, 87 and 90.
Thus, the number of times, the children would say ‘idli’ is 30.

(ii) The multiples of 5 from 1 to 90 are 5, 10,15, 20, 25, 30, 35,40, 45, 50, 55,60, 65, 70, 75, 80, 85 and 90.
Thus, the number of times, the children would say ‘vada’ is 18.

(iii) The multiples of 15 are 15, 30,45,60, 75 and 90. Thus, the number of times, the children would say ‘idli-vada’ is 6.

Question 3.
What if the game was played till 900? How would your answers change?

Solution:
If the game was played till 900 the numbers which are multiple of 3 are 300 and the multiples of 5 are 180.

Question 4.
Is this figure somehow related to the ‘idli-vada’ game?

Hint: Imagine playing the game till 30. Draw the figure if the game is played till 60.

Solution:

Yes the given figure is related to the ‘idli-vada’ game. The numbers for ‘idli’ are given in the left circle, the numbers for ‘vada’ are given in the right circle and the numbers common in both the circles are for’idli-vada’.

Question 1.
Find out other such numbers that are multiples of both 3 and 5. These numbers are called ___ (Page 107)

Solution:
Numbers that are multiples of both 3 and 5 are called common multiples. The smallest common multiple of 3 and 5 is 15, and other such numbers include 30, 45, 60, and so on.

Question 2.
Which of the following could be the other number: 2, 3, 5, 8,10? (Page 109)
Answer:
2

Question 3.
What jump size can reach both 15 and 30? There are multiple jump sizes possible. Try to find them all. f (Page 110)

Solution:
To find the jump sizes that allow Jumpy to land on both 15 and 30, you need to determine the common factors of these two numbers. Here’s how you can find these common jump sizes:

Factors of 15:

• 15 can be factored into: 15 = 3 × 5

• The factors of 15 are: 1, 3, 5, 15

Factors of 30:

• 30 can be factored into: 30 = 2 × 3 × 5

• The factors of 30 are: 1, 2, 3, 5, 6, 10, 15, 30

The common factors between these two lists are: 1, 3, 5, 15. So, the jump sizes that will allow Jumpy to land on both 15 and 30 are the common factors of 15 and 30.
Therefore, the jump sizes that will enable Jumpy to land on both 15 and 30 are: 1, 3, 5, 15

Question 4.
Which numbers are both shaded and circled? What are these numbers called? (Page 110)

Solution:
36,48,60 are both shaded and circled. These numbers are called common multiples.

5.1 Common Multiples and Common Factors Figure it Out (Page No. 110 – 111)

Question 1.
Find all multiples of 40 that lie between 310 and 410.

Solution:
Multiples of 40 that lie between 310 and 410 are 320, 360 and 400.

Question 2.
Who am I?
(a) I am a number less than 40. One of my factors is 7. The sum of my digits is 8.
(b) I am a number less than 100. Two of my factors are 3 and 5. One of my digits is 1 more than the other.

Solution:
(a) The numbers less than 40 whose one of the factor is 7 are 7,14, 21, 28, 35. Out of these numbers the sum of digits of 35 is 8.
Hence, the number less than 40 whose one of the factors is 7 and sum of digits equals to 8 is 35.

(b) The number less than 100 whose two factors 3 and 5 are 15, 30, 45,60, 75, 90. Out of these numbers, the number 15 has one of digits is 1 more than the other.
Hencd, the number less than 100 whose two factors are 3 and 5 and one of digits is 1 more than the other is 15.

Question 3.
A number for which the sum of all its factors is equal to twice the number is called a perfect number. The number 28 is perfect. Its factors are 1, 2, 4, 7, 14 and 28. Their sum is 56 which is twice 28. Find a perfect number between 1 and 10.

Solution:
A perfect number between 1 and 10 is 6. Its factors are 1, 2, 3, and 6 and the sum of all factors is 12 which is twice of 6.

Question 4.
Find the common factors of:
(a) 20 and 28
(b) 35 and 50
(c) 4, 8 and 12
(d) 5, 15 and 25

Solution:
(a) Factors of 20 are 1, 2, 4, 5, 10, 20.
Factors of 28 are 1, 2, 4, 7, 14, 28.
So the factors that are common to both of the given numbers are 1, 2, and 4.
Therefore, the common factors are 1, 2, and 4.

(b) Factors of 35 are 1, 5, 7, 35
Factors of 50 are 1, 2, 5, 10, 25, 50
So, the factors which are common to both of the given numbers are 1 and 5.
Therefore, the common factors are 1 and 5.

(c) Factors of 4 are 1, 2, 4
Factors of 8 are 1, 2, 4, 8
Factors of 12 are 1, 2, 3, 4, 6, 12.
So, the factors that are common to all of the given numbers are 1, 1, and 4.
Therefore, the common factors are 1, 2, and 4.

(d) Factors of 5 are 1 and 5.
Factors of 15 are 1, 3, 5 and 15.
Factors of 25 are 1, 5, and 25.
So, the factors which are common to all of the given numbers are 1 and 5.
Therefore, the common factors are 1 and 5.

Question 5.
Find any three numbers that are multiples of 25 but not multiples of 50.

Solution:
Numbers that are multiples of 25 are 25, 50, 75, 100, 125, 150, 175, …
Numbers that are multiples of 50 are 50, 100, 150, 200, 250,300,…
Hence, the numbers that are multiples of 25 but not multiples of 50 are 25, 75, 125, 175,…

Question 6.
Anshu and his friends play the ‘idli-vada ’ game with two numbers, which are both smaller than 10. The first time anybody says ‘idli-vada’ is after the number 50. What could the two numbers be which are assigned ‘idli’and ‘vada’?

Solution:
If ‘idli-vada’ is said after number 50 it means that the least common multiple (LCM) of the two numbers must be slightly greater than 50. The LCM of 6 and 9 is 54,’.which is the first common multiple after 50, making 6 and 9 the possible numbers. Hence the two numbers could be 6 and 9.

Question 7.
In the treasure hunting game, Grumpy has kept treasures on 28 and 70. What jump sizes will land on both the numbers?

Solution:
Factors of 28 = 1, 2, 4, 7, 14, 28
Factors of 70 = 1,2, 5, 7, 10, 14, 35, 70
Common factors are 1, 2, 7 and 14
Hence jump sizes which will land at both 28 and 70 are 1, 2, 7 and 14.

Question 8.
In the diagram below,
Guna has erased all the numbers except the common multiples. Find out what those numbers could be and fill in the missing numbers in the empty regions.

Solution:
A pair of possible number
whose common multiples are 24, 48 and 72 would be 6 and 8.
Multiples of 6 = 6, 12, 18, 24, 30, 36, 42, 48, 54, 60, 66, 72,…
Multiples of 8 = 8, 16, 24, 32, 40, 48, 56, 64, 72, …

Other possible pairs are 3 and 8; 8 and 12.

Question 9.
Find the smallest number that is a multiple of all the numbers from 1 to 10 except for 7.

Solution:
The smallest number that is a multiple of all the numbers from 1 to 10 except for 7 will be the least common multiple of these numbers.

2 × 2 × 3 × 5 × 2 × 3 = 360
Thus, the smallest number that is a multiple of all the numbers from 1 to 10 except for 7 will be 360.

Question 10.
Find the smallest number that is a multiple of all the numbers from 1 to 10.

Solution:
To find the shiallesf number that is a multiple of all
numbers from 1 to 10, we need to determine LCM of the numbers from 1 to 10.
Prime factorisation of numbers from 1 to 10.
1 = 1
2 = 2
3 = 3
4 = 2 × 2
5 = 5
6 = 2 × 3
7 = 7
8 = 2 × 2 × 2
9 = 3 × 3
10 = 2 × 5
LCM (1 to 10) = 23 × 32 × 5 × 7
= 2520
Thus, the required smallest number is 2520.

Read More: Addition and Subtraction of Time

5.2 Prime Numbers Figure it Out (Page No. 114 – 115)

Question 1.
We see that 2 is a prime and also an even number. Is there any other even prime?

Solution:
No, 2 is the only even prime number.

Question 2.
Look at the list of primes till 100. What is the smallest difference between two successive primes? What is the largest difference?

Solution:
Prime numbers upto 100 are 2, 3, 5, 7,11,13,17,19, 23, 29, 31, 37,41,43,47, 53, 59, 61, 67, 71, 73, 79, 83, 89 and 97
Difference 3 – 2 = 1 and 97 – 89 = 8
Smallest difference: The smallest difference between successive prime is 1 (i.e. between 2 and 3).
Largest difference: The largest difference between successive prime is 8 (i.e. between 89 and 97).

Question 3.
Are there an equal number of primes occurring in every row in the table? Which decades have the least number of primes? Which have the most number of primes?

Solution:
No, there are not an equal number of primes occurring in every row in the table. The last row has the least number of primes and the first and second rows have the maximum prime numbers.

Question 4.
Which of the following numbers is prime?
23, 51, 37, 26

Solution:

23 and 37 are the prime numbers.

Question 5.
Write three pairs of prime numbers less than 20 whose sum is a multiple of 5.

Solution:
Three pairs of prime numbers less than 20 whose sum is a multiple of 5 are: (2, 3), (2, 13) and (7, 13).

Question 6.
The numbers 13 and 31 are prime numbers. Both these numbers have same digits 1 and 3. Find such pairs of prime numbers up to 100.

Solution:
The valid pairs of prime numbers up to 100 that consist of the same digits are: (13, 31), (17, 71), (37, 73) and (79, 97).

Question 7.
Find seven consecutive composite numbers between 1 and 100.

Solution:
Seven consecutive composite numbers between 1 and 100 are 90, 91, 92, 93, 94, 95 and 96.

Question 8.
Twin primes are pairs of primes having a difference of 2. For example, 3 and 5 are twin primes. So are 17 and 19. Find the other twin primes between 1 and 100.

Solution:
Twin primes between 1 and 100 are
(3, 5) → 5 – 3 = 2
(5, 7) → 7 – 5 = 2
(11, 13) → 13 – 11 = 2
(17, 19) → 19 – 17 = 2

(29, 31) → 31 – 29 = 2

(41, 43) → 43 – 41 = 2

(59, 61) → 61 – 59 = 2

(71, 73) → 73 – 71 = 2

Question 9.
Identify whether each statement is true or false. Explain.
(a) There is no prime number whose units digit is 4.
(b) A product of primes can also be prime.
(c) Prime numbers do not have any factors.
(d) All even numbers are composite numbers.
(e) 2 is a prime and so is the next number, 3. For every other prime, the next number is composite.

Solution:
(a) (True)
Any number ending in 4 is divisible by 2, making it an even number.
Except for the number 2, no even number is prime.

(b) (False)
No, a product of primes cannot be a prime.
e.g. 2 and 3 are prime numbers but 2×3 = 6, which is not a prime number.
When you multiply two or more primes, you get a composite number, which has divisors other than just 1 and itself.

(c) (False)
Prime numbers have factors but only two
i. e. 1 and the number itself.
e.g. Prime number 5 has only two factors 1 and 5.

(d) (False)
No, all even numbers are not composite numbers, e.g. 2 is an even prime number.

(e) (True)
2 is a prime followed by a prime number 3.
For every other prime, the next number is composite, e.g. 5 followed by 6, which is composite.
11 followed by 12, which is composite.
13 followed by 14, which is composite.

Question 10.
Which of the following numbers is the product of exactly three distinct prime numbers: 45, 60, 91, 105, 330?

Solution:
Here, 45 = 3 × 3 × 5 (2 distinct primes)
60 = 2 × 2 × 3 × 5(3 distinct primes)
91 = 7 × 13 (2 distinct primes)
105 = 3 × 5 × 7 (3 distinct primes)
330 = 2 × 3 × 5 × 11 (4 distinct primes)
Number 105 is the product of exactly three distinct prime numbers i.e. 3 × 5 × 7.

Question 11.
How many three-digit prime numbers can you make using each of 2, 4 and 5 once?

Solution:
The possible numbers formed by 2,4 and 5 are: 245, 254, 425, 452, 524, 542.
These all are composite numbers not prime numbers. So, no prime number can be made using each of 2,4 and 5 once.

Question 12.
Observe that 3 is a prime number, and 2 × 3 + 1 = 7 is also a prime. Are there other primes for which doubling and adding 1 gives another prime? Find at least five such examples.

Solution:
The five prime numbers for which doubling and adding 1 gives another prime are:

• 2 (since 2 × 2 + 1 = 5)

• 3 (since 2 × 3 + 1 = 7)

• 5 (since 2 × 5 + 1 = 11)

• 11 (since 2 × 11 + 1 = 23)

• 23 (since 2 × 23 + 1 = 47)

Question 1.
How many prime numbers are there from 21 to 30? (Page 113)

Solution:
In total, there are 2 prime numbers between 21 and 30.
They are 23 and 29.

Question 2.
How many composite numbers are there from 21 to 30? (Page 113)

Solution:
Total number of composite numbers from 21 to 30 is 8.
They are 21,22,24,25, 26, 27, 28, 30.

Read More: CBSE Class 8 Maths Formulas

5.3 Co-prime Numbers for Safekeeping Treasures Figure it Out (Page No. 120)

Question 1.
Find the prime factorisations of the following numbers: 64, 104, 105, 243, 320, 141, 1728, 729, 1024, 1331, 1000.
Solution:

Question 2.
The prime factorisation of a number has one 2, two 3s, and one 11. What is the number?

Solution:
The prime factorisation of a number has one 2, two 3s, and one 11.
Therefore, required number = 2 × 3 × 3 × 11 = 198

Question 3.
Find three prime numbers, all less than 30, whose product is 1955.

Solution:

5 × 17 × 23 = 1955.

Thus, 5, 17 and 23 are three prime numbers 17 less than 30 whose product is 1955.

Question 4.
Find the prime factorisation of these numbers without multiplying first
(a) 56 × 25
(b) 108 × 75
(c) 1000 × 81

Solution:
(a) Given number is 56 × 25.
Prime factorisation of 56 = 2 × 2 × 2 × 7
Prime factorisation of 25 = 5 × 5
∴ Prime factorisation of 56 × 25
= 2 × 2 × 2 × 7 × 5 × 5 = 2 × 2 × 2 × 5 × 5 × 7

(b) Given number is 108 × 75.
Prime factorisation of 108 = 2 × 2 × 3 × 3 × 3
Prime factorisation of 75 = 3 × 5 × 5
∴ Prime factorisation of 108 × 75
= 2 × 2 × 3 × 3 × 3 × 3 × 5 × 5

(c) Given number is 1000 × 81.
Prime factorisation of 1000 = 2 × 2 × 2 × 5 × 5 × 5
Prime factorisation of 81 = 3 × 3 × 3 × 3
∴ Prime factorisation of 1000 × 81
= 2 × 2 × 2 × 5 × 5 × 5 × 3 × 3 × 3 × 3
= 2 × 2 × 2 × 3 × 3 × 3 × 3 × 5 × 5 × 5

Question 5.
What is the smallest number whose prime factorisation has:
(a) three different prime numbers?
(b) four different prime numbers?

Solution:
(a) The smallest three prime numbers are 2, 3 and 5.
Thus, the smallest number with exactly three different prime factors = 2 × 3 × 5 = 30

(b) The smallest four prime numbers are 2,3,5 and 7.
Thus, the smallest number with exactly four different prime factors = 2 × 3 × 5 × 7 = 210

Intext Questions

Question 1.
Which of the following pairs of numbers are co-prime? (Page 116)
(a) 18 and 35
(b) 15 and 37
(c) 30 and 415

Answer:
(a) Here factors of 18 = 1 × 2 × 3 × 3 and factors of 35 = 1 × 5 × 7

No common factor other than 1.

Hence 18 and 35 are co-prime numbers.

(b) We have factors of 15 = 1 × 3 × 5 and factors of 37 = 1 × 37
No common factor other than 1.
Hence 15 and 37 are co-prime numbers.

(c) Given numbers are 30 and 415 Here factors of 30 = 1 × 2 × 3 × 5 and factors of 415 = 5 × 83
Clearly 5 is a common factor of 30 and 415.
Hence 30 and 415 are not co-prime numbers.

5.4 Prime Factorisation Figure it Out (Page No. 122)

Question 1.
Are the following pairs of numbers co-prime? Guess first and then use prime factorization to verify your answer.
(a) 30 and 45
(b) 57 and 85
(c) 121 and 1331
(d) 343 and 216

Solution:
(a) Factors of 30 and 45:
30 = 2 × 3 × 5,
45 = 3 × 3 × 5
Common factors: 3 × 5 = 15, hence 30 and 45 are not a pair of co-prime numbers.

(b) Factors of 57 and 85:
57 = 3 × 19,
85 = 5 × 17
No common factors other than 1, hence 57 and 85 are a pair of co-prime numbers.

(c) Factors of 121 and 1331:
121 = 11 × 11,
1331 = 11×11×11
Common factors: 11 × 11 = 121, hence 121 and 1331 are not a pair of co-prime numbers.

(d) Factors of 343 and 216:
343 = 7 × 7 × 7,
216 = 2 × 2 × 2 × 3 × 3 × 3
No common factors other than 1, hence 343 and 216 are a pair of co-prime numbers.

Question 2.
Is the first number divisible by the second? Use prime factorization.
(a) 225 and 27
(b) 96 and 24
(c) 343 and 17
(d) 999 and 99.

Solution:
(a) The prime factorization of 225 is 3 × 3 × 5 × 5 and 27 is 3 × 3 × 3.
No, 3 occurs twice in the prime factorization of 225 but thrice in the prime factorization of 27.
So, 225 is not divisible by 27.

(b) The prime factorization of 96 is 2 × 2 × 2 × 2 × 2 × 3 and 24 is 2 × 2 × 2 × 3.
Yes, all the prime factors of 24 occur in the prime factorization of 96.
So, 96 is divisible by 24.

(c) The prime factorization of 343 is 7 × 7 × 7 and 17 is 1 × 7.
No, all the prime factors of 343 are different from the prime factors of 17.
So, 343 is not divisible by 17.

(d) The prime factorization of 999 is 3 × 3 × 3 × 37 and 99 is 3 × 3 × 11
No, 3 occurs thrice in the prime factorization of 999 but twice in the prime factorization of 99.
So, 999 is not divisible by 99.

Question 3.
The first number has prime factorization 2 × 3 × 7 and the second number has prime factorization 3 × 7 × 11. Are they co-prime? Does one of them divide the other?

Solution:
The numbers share the common factors 3 and 7. So they are not co-prime since neither number contains all the factors of the other, neither can divide the other.

Question 4.
Guna says, “Any two prime numbers are co-prime”. Is he right?

Solution:

Yes, Guna is right. Any two prime numbers are co-prime as they do not have common factor other than 1 which means they are always co-prime. For example, 2 and 3, 5 and 7, 11 and 13.

5.5 Divisibility Tests Figure it Out (Page No. 125-126)

Question 1.
2024 is a leap year (as February has 29 days). Leap years occur in the years that are multiples of 4, except for those years that are evenly divisible by 100 but not 400.
(a) From the year you were bom till now, which years were leap years?
(b) From the year 2024 till 2099, how many leap years are there?

Solution:
(a) Do it yourself.
(b) We know that leap years occur in the years that are multiples of 4, except century years.
Therefore, multiples of 4 after 2024, will be 2028, 2032, 2036, 2040, 2044, 2048, 2052, 2056, 2060, 2064, 2068, 2072, 2076, 2080, 2084, 2088, 2092, and 2096, till 2099.
Therefore, there will be 19 leap years from the year 2024 till 2099.

Question 2.
Find the largest and smallest 4-digit numbers that are divisible by 4 and are also palindromes.

Solution:
The smallest 4-digit palindrome is 1001.
Checking divisibility by 4 1001 (not divisible by 4)
1111 (not divisible by 4) 1221 (not divisible by 4)
1331 (not divisible by 4)
1441 (not divisible by 4)
1551 (not divisible by 4)

1661 (not divisible by 4)

1771 (not divisible by 4)

1881 (not divisible by 4)

1991 (not divisible by 4)

2002 (not divisible by 4)

2112 (divisible by 4)

Therefore, the smallest 4-digit palindromic number divisible by 4 is 2112.

The largest 4-digit palindrome is 9999.
Checking divisibility by 4 9999 (not divisible by 4)
9889 (not divisible by 4)
9779 (not divisible by 4)
9669 (not divisible by 4)
9559 (not divisible by 4)
9449 (not divisible by 4)
9339 (not divisible by 4)
9229 (not divisible by 4)
9119 (not divisible by 4)
9009 (not divisible by 4)
8998 (not divisible by 4)
8888 (divisible by 4)
Therefore, the largest 4-digit palindromic number divisible by 4 is 8888.

Question 3.
Explore and find out if each statement is always true, sometimes true or never true. You can give e×amples to support your reasoning.
(a) Sum of two even numbers gives a multiple of 4.

Solution:
Sometimes true. Sum of any two even numbers is not always divisible by 4. For e×ample, 6 + 4 = 10 which is not divisible by 4 whereas 2+2 = 4 which is divisible of 4.

(b) Sum of two odd numbers gives a multiple of 4.

Solution:
Sometimes true. Sum of two odd numbers can indeed be even but not necessarily a multiple of 4. For e×ample, 1 + 5 = 6 which is not a multiple of 4 whereas 1+3 = 4, which is a multiple of 4. Similarly 7 + 5 = 12, which is a multiple of 4.

Question 4.
Find the remainders obtained when each of the following numbers are divided by
(i) 10, (ii) 5, (iii) 2.
78, 99, 173, 572, 980, 1111,2345

Solution:
(i) When divided by 10:
78 ÷ 10: Q = 7 and R = 8;
99 ÷ 10: Q = 9 and R = 9;
173÷ 10: Q = 17 and R = 3;
572 ÷ 10: Q = 57 and R = 2;
980 ÷ 10: Q = 98 andR = 0;
1111 ÷ 10: Q = 111 and R = 1;
2345 ÷ 10: Q = 234 and R = 5
Note: To find the remainder when dividing a number by 10, simply check how much the last digit exceeds ‘O’.

(ii) When divided by 5
78 ÷ 5: Q= 15 andR = 3;
99 ÷ 5: Q = 19 and R = 4;
173 ÷ 5: Q = 34 andR = 3;
572 ÷ 5: Q = 114 and R = 2;
980 ÷ 5: Q= 196 and R = 0;
1111 ÷ 5: Q = 222 and R= 1;
2345 ÷ 5: Q = 469 and R = 0
Note: To find the remainder when dividing by 5, check how much the last digit exceeds the nearest ‘0’ or ‘5’.

(iii) When divided by 2
78 ÷ 2: Q = 39 and R = 0;
99 ÷ 2: Q = 49 and R = 1;
173 ÷ 2; Q = 86 and R = 1;
572 ÷ 2; Q = 286 and R = 0;
980 ÷ 2; Q = 490 and R = 0;
1111 ÷ 2: Q = 555 and R = 1;
2345 ÷ 2: Q= 1172 and R = 1.
Note: To find the remainder when dividing by 2, check if the given digits is even (remainder ‘0’) or odd (remainder ‘1’)-

Question 5.
The teacher asked if 14560 is divisible by all of 2, 4, 5, 8 and 10. Guna checked for divisibility of 14560 by only two of these numbers and then declared that it was also divisible by all of them. What could those two numbers be?

Solution:
If a number is divisible by 8, it will automatically be divisible by 4.
If a number is divisible by 10, it is also divisible by 2 and 5. Therefore, checking divisibility by 8 and 10 confirms divisibility by all other numbers (2, 4, 5).
Thus, the pair of numbers that Guna could check to determine that 14560 is divisible by all of 2, 4, 5, 8, and 10 is: 8 and 5.

Question 6.
Which of the following numbers are divisible by all of 2,4,5,8 and 10:572,2352,5600,6000,77622160?

Solution:
When a number is divisible by 8 and 10 it is also divisible by all the given numbers.
The numbers that are divisible by 2, 4, 5, 8, and 10 are 5600, 600,0 and 77622160.

Question 7.
Write two numbers whose product is 10000. The two numbers should not have 0 as the units digit.

Solution:
We need to write factors of 10000.
10000= 10 × 10 × 10 × 10
= 2 × 5 × 2 × 5 × 2 × 5 × 2 × 5
So, 2 × 2 × 2 × 2 = 16 and 5 × 5 × 5 × 5 = 625.
Hence, 16 and 625 are the two numbers whose product is 10000.

Intext Questions (Page 124)

Question 1.
What are all the multiples of 2 between 399 and 411?

Solution:
The multiples of 2 between 399 and 411 are 400, 402, 404, 406, 408,and 410.

Question 2.
Find numbers between 330 and 340 that are divisible by 4. Also, find numbers between 1730 and 1740, and 2Q30 and 2040, that are divisible by 4. What do you observe?

Solution:
The numbers between 330 and 340 that are divisible by 4 are 332, 336, and 340. The numbers between 1730 and 1740 that are divisible by 4 are 1732, 1736, and 1740. The numbers between 2030 and 2040 that are divisible by 4 are 2032, 2036, and 2040.

Question 3.
Is 8536 divisible by 4?

Solution:
Since the last two digits of 8536 (which is 36) are divisible by 4, the original number 8536 is also divisible by 4.

Question 4.
Consider these statements:
(a) Only the last two digits matter when deciding if a given number is divisible by 4.
(b) If the number formed by the last two digits is divisible by 4, then the original number is divisible by 4.
(c) If the original number is divisible by 4, then the number formed by the last two digits is divisible by 4.
Do you agree? Why or why not?

Solution:
(a) Yes, that’s correct. When determining if a number is divisible by 4, only the last two digits of the number matter. This is because 100 is divisible by 4, so the divisibility rule for 4 focuses on whether the number formed by the last two digits is divisible by 4.
(b) Yes, that’s correct. If the number formed by the last two digits of a given number is divisible by 4, then the original number is also divisible by 4.
(c) Yes, that’s correct. If the original number is divisible by 4, the last two digits of the number will indeed be divisible by 4.

Read More: Successor and Predecessor in Maths

Breakdown of Chapter 5 Class 6 Maths Prime Time

The "Prime Time" chapter is designed to help students become more comfortable with numbers by teaching them certain ideas. Chapter 5 class 6 maths Prime Time has a lot of different types of problems and fundamental concepts. 

1. Factors and Multiples
   A factor is a number that divides another number without leaving a residue. On the other hand, a multiple is the result of multiplying an integer by any whole number.

• Example from Reference: To discover the factors of 12, we seek for pairs of numbers that add up to 12: (1, 12), (2, 6), and (3, 4). So, the numbers that work are 1, 2, 3, 4, 6, and 12.

• Note about numbers: 1 is a factor of every number, and every number is a factor of itself.

2. Numbers that are prime and numbers that are composite
   This chapter is mostly on the difference between prime and composite numbers.

• Prime Numbers: Numbers like 2, 3, 5, 7, and 11 are prime numbers because they only have two factors. 1 and the number itself.

• Composite Numbers: Composite numbers are numbers that have more than two factors, such as 4, 6, 8, 9, and 10.

• Special Case: The number 1 is not a prime number or a composite number. The only even prime number is 2, which is also the smallest prime number.

3. Rules for Divisibility
   Divisibility rules are tests that tell you if one number can be divided by another without actually doing the division. According to the NCERT answers:

• Divisibility by 2: A number is divisible by 2 if its last digit is 0, 2, 4, 6, or 8.

• Divisibility by 3: A number is divisible by 3 if the sum of its digits is a multiple of 3.

• Divisibility by 5: A number is divisible by 5 if its last digit is 0 or 5.

• Divisibility by 10: A number is divisible by 10 if its last digit is 0.

4. Finding the prime factors
   This means writing a composite number as the product of its prime elements. The factor tree method is the most prevalent way to solve problems in chapter 5 in class 6 maths. For example, you may write the prime factorisation of 24 as  23 × 3. 

5. HCF and common factors
   "Common factors" are the factors that two or more integers have in common. The Highest Common Factor (HCF) is the biggest of these.
   
   Problem Scenario: Find the HCF of 12 and 18.

• Factors of 12: 1, 2, 3, 4, 6, 12.

• Factors of 18: 1, 2, 3, 6, 9, 18.

• Common Factors: 1, 2, 3, 6.

• HCF: 6.

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