---
title: "NCERT Solutions for Class 6 Maths Chapter 5 Prime Time (2026-27)"
url: https://www.swavid.com/maths/class/6/chapter/prime-time/ncert-solutions
dateModified: 2026-10-07T14:51:01+00:00
---

# NCERT Solutions for Class 6 Maths Chapter 5 Prime Time (2026-27)

This chapter's questions cover fundamental concepts in number theory such as multiples, factors, prime and composite numbers, prime factorisation, divisibility tests, and co-prime numbers. Students explore these ideas through problem-solving and reasoning tasks.

Free PDF (19 pages): https://www.swavid.com/api/seo/pdf/ncert/maths/class-6/swavid-ncert-solutions-class-6-maths-chapter-5-prime-time-86020e29d9.pdf

## Figure it Out

### Question 1

*2 marks · Very short answer*

At what number is ‘idli-vada’ said for the 10th time?

**Solution**

1. The phrase 'idli-vada' is said at numbers that are common multiples of 3 and 5.
2. The common multiples of 3 and 5 are multiples of their LCM, which is 15.
3. The 10th time 'idli-vada' is said will be at the 10th multiple of 15, which is $15 \times 10 = 150$.

**Answer:** 150

> Common mistake: Counting individual multiples incorrectly without using the least common multiple.

### Question 2

*3 marks · Short answer*

If the game is played for the numbers from 1 to 90, find out:
a. How many times would the children say ‘idli’ (including the times they say ‘idli-vada’)?
b. How many times would the children say ‘vada’ (including the times they say ‘idli-vada’)?
c. How many times would the children say ‘idli-vada’?

**Part a**

1. The numbers from 1 to 90 that are multiples of $3$ are counted by dividing $90$ by $3$.
2. There are $30$ multiples of $3$ between 1 and 90.
3. So, the children say 'idli' $30$ times.

Answer a: $30$ times

**Part b**

1. The numbers from 1 to 90 that are multiples of $5$ are counted by dividing $90$ by $5$.
2. There are $18$ multiples of $5$ between 1 and 90.
3. So, the children say 'vada' $18$ times.

Answer b: $18$ times

**Part c**

1. The numbers that are both multiples of $3$ and $5$ are multiples of their product $15$.
2. The multiples of $15$ up to 90 are $15, 30, 45, 60, 75, \text{ and } 90$.
3. There are $6$ such numbers, so the children say 'idli-vada' $6$ times.

Answer c: $6$ times

**Answer:** The children say 'idli' 30 times, 'vada' 18 times, and 'idli-vada' 6 times.

> Common mistake: Forgetting to include 'idli-vada' counts when counting 'idli' or 'vada' totals.

### Question 3

*3 marks · Short answer*

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

**Solution**

1. When the game is played up to 900, the number of multiples of 3 is calculated by dividing 900 by 3, which gives 300 times for 'idli'.
2. The number of multiples of 5 is calculated by dividing 900 by 5, which gives 180 times for 'vada'.
3. The number of common multiples of 3 and 5 (multiples of 15) is calculated by dividing 900 by 15, which gives 60 times for 'idli-vada'.

**Answer:** 'Idli' will be said 300 times, 'vada' will be said 180 times, and 'idli-vada' will be said 60 times.

> Common mistake: Calculating up to 90 instead of 900.

### Question 4

*3 marks · Short answer*

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**

1. The figure is a Venn diagram showing multiples of 3 and 5 using circles.
2. The region where the two circles overlap represents the numbers that are multiples of both 3 and 5.
3. These common numbers in the overlapping region represent the numbers when players say 'idli-vada'.

**Answer:** Yes, the common numbers in the intersecting region represent the numbers when to say 'idli-vada'.

> Common mistake: Failing to connect the overlapping regions of a Venn diagram to common multiples.

## Figure it Out

### Question 1

*2 marks · Very short answer*

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

**Solution**

1. The multiples of 40 are obtained by multiplying 40 by 1, 2, 3, etc., giving 40, 80, 120, 160, 200, 240, 280, 320, 360, 400, 440, and so on.
2. The multiples of 40 that lie between 310 and 410 are 320, 360, and 400.

**Answer:** 320, 360, 400

> Common mistake: Including multiples outside the given range, such as 280 or 440.

### Question 2

*3 marks · Short answer*

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.

**Part a (1 mark)**

1. Multiples of 7 less than 40 are 7, 14, 21, 28, and 35.
2. Among these, the number whose digits add up to 8 is $3 + 5 = 8$.
3. Therefore, the number is 35.

Answer a: 35

**Part b (2 marks)**

1. Numbers less than 100 that have both 3 and 5 as factors are multiples of their common multiple, 15.
2. The multiples of 15 less than 100 are 15, 30, 45, 60, 75, and 90.
3. For 45, the digits are 4 and 5, and one digit is 1 more than the other ($5 - 4 = 1$).
4. Therefore, the number is 45.

Answer b: 45

**Answer:** a. 35, b. 45

> Common mistake: Forgetting to check the digit conditions given in the second part.

### Question 3

*2 marks · Very short answer*

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 a perfect number. 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**

1. The factors of 6 are 1, 2, 3, and 6.
2. The sum of all its factors is $1 + 2 + 3 + 6 = 12$, which is twice of 6.
3. Therefore, 6 is a perfect number between 1 and 10.

**Answer:** 6

> Common mistake: Confusing prime numbers with perfect numbers.

### Question 4

*3 marks · Short answer*

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

**Part a**

1. The factors of 20 are 1, 2, 4, 5, 10, 20 and the factors of 28 are 1, 2, 4, 7, 14, 28.
2. The common factors of 20 and 28 are 1, 2, and 4.

Answer a: 1, 2, 4

**Part b (1 mark)**

1. The factors of 35 are 1, 5, 7, 35 and the factors of 50 are 1, 2, 5, 10, 25, 50.
2. The common factors of 35 and 50 are 1 and 5.

Answer b: 1, 5

**Part c (1 mark)**

1. The 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.
2. The common factors of 4, 8 and 12 are 1, 2, and 4.

Answer c: 1, 2, 4

**Part d (1 mark)**

1. The factors of 5 are 1, 5; factors of 15 are 1, 3, 5, 15; factors of 25 are 1, 5, 25.
2. The common factors of 5, 15 and 25 are 1 and 5.

Answer d: 1, 5

**Answer:** a. 1, 2, 4; b. 1, 5; c. 1, 2, 4; d. 1, 5

> Common mistake: Missing out 1 as a common factor.

### Question 5

*2 marks · Very short answer*

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

**Solution**

1. Multiples of 25 are numbers like 25, 50, 75, 100, 125, 150, 175, and so on.
2. To ensure they are not multiples of 50, we select the odd multiples of 25.
3. Three such numbers are 25, 75, and 125.

**Answer:** 25, 75, 125

> Common mistake: Including numbers like 50 or 100 which are multiples of 50.

### Question 6

*3 marks · Short answer*

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**

1. The first time 'idli-vada' is said is at the LCM (first common multiple) of the two numbers, which is greater than 50.
2. If the two numbers are 7 and 8, their first common multiple is 56, which is greater than 50.
3. If the two numbers are 8 and 9, their first common multiple is 72, which is greater than 50.
4. Therefore, the two numbers could be 7 and 8, or 8 and 9.

**Answer:** 7 and 8 (or 8 and 9)

> Common mistake: Choosing numbers whose common multiple is less than or equal to 50, such as 6 and 8.

### Question 7

*2 marks · Very short answer*

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

**Solution**

1. The jump sizes that land on both numbers are the common factors of 28 and 70.
2. The common factors of 28 and 70 are 1, 2, 7, and 14.

**Answer:** The jump sizes are 1, 2, 7, or 14.

> Common mistake: Listing only some factors instead of all common factors.

### Question 8

*3 marks · Short answer*

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**

1. The common multiples given are 24, 48, and 72, which are the multiples of 24.
2. The two numbers whose common multiples are 24, 48, and 72 can be 8 and 12, or 3 and 8.
3. Filling the regions with respective multiples of the two numbers gives the required Venn diagram.

**Answer:** Multiples of 8 and 12 with common multiples 24, 48, 72.

> Common mistake: Confusing factors with multiples.

### Question 9

*2 marks · Very short answer*

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

**Solution**

1. The prime factorisations of numbers from 1 to 10 except 7 are: $2 = 2$, $3 = 3$, $4 = 2^2$, $5 = 5$, $6 = 2 \times 3$, $8 = 2^3$, $9 = 3^2$, $10 = 2 \times 5$.
2. Taking the highest power of each prime factor ($2^3$, $3^2$, and $5$), the product is $8 \times 9 \times 5 = 360$.

**Answer:** 360

> Common mistake: Including 7 in the product or missing the highest power of prime factors.

### Question 10

*2 marks · Very short answer*

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

**Solution**

1. The prime factorisations of numbers from 1 to 10 are: $2 = 2$, $3 = 3$, $4 = 2^2$, $5 = 5$, $6 = 2 \times 3$, $7 = 7$, $8 = 2^3$, $9 = 3^2$, $10 = 2 \times 5$.
2. Taking the highest power of each prime factor ($2^3$, $3^2$, $5$, and $7$), the product is $8 \times 9 \times 5 \times 7 = 2520$.

**Answer:** 2520

> Common mistake: Multiplying all numbers directly without considering prime factorisation and highest powers.

## Figure it Out

### Question 1

*2 marks · Very short answer*

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

**Solution**

1. A prime number has only two factors, 1 and itself.
2. Any other even number has 2 as a factor in addition to 1 and itself, making it composite, so 2 is the only even prime number.

**Answer:** No, 2 is the only even prime number.

> Common mistake: Stating that there are other even primes like 4 or 6 by confusing prime and composite numbers.

### Question 2

*3 marks · Short answer*

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

**Solution**

1. List the prime numbers up to 100 from the Sieve of Eratosthenes table in the textbook.
2. Observe the differences between successive prime numbers, noting that 2 and 3 have a difference of $3 - 2 = 1$, while all other successive primes have a difference of at least 2.
3. Find the largest difference between successive primes in the list up to 100, which is between 89 and 97 with a difference of $97 - 89 = 8$.

**Answer:** The smallest difference between two successive primes is 1 (between 2 and 3), and the largest difference is 8 (between 89 and 97).

> Common mistake: Stating that the smallest difference between successive primes is 2 by ignoring the prime pair 2 and 3.

### Question 3

*3 marks · Short answer*

Are there an equal number of primes occurring in every row in the table on the previous page? Which decades have the least number of primes? Which have the most number of primes?

**Solution**

1. Observe the table of prime numbers up to 100 on page 113.
2. Count the prime numbers in each row or decade, finding that they are not equal in number.
3. Identify that the decades 1 to 10 and 11 to 20 have the most primes (4 primes each) while the decade 91 to 100 has the least (1 prime).

**Answer:** No, the number of primes in each row is not equal. The decades 1 to 10 and 11 to 20 have the most primes (4 primes each) and the decade 91 to 100 has the least number of primes (1 prime).

> Common mistake: Counting the primes incorrectly by missing primes like 2 or 11 in the respective decades.

### Question 4

*2 marks · Very short answer*

Which of the following numbers are prime: 23, 51, 37, 26?

**Solution**

1. Check the factors of each number: 23 has factors 1 and 23; 51 can be divided by 3 ($3 \times 17 = 51$); 37 has factors 1 and 37; 26 is even and divisible by 2.
2. Therefore, only numbers with exactly two factors are prime.

**Answer:** 23 and 37 are prime numbers.

> Common mistake: Mistaking 51 for a prime number, forgetting that it is divisible by 3.

### Question 5

*3 marks · Short answer*

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

**Solution**

1. List the prime numbers less than 20: 2, 3, 5, 7, 11, 13, 17, 19.
2. Select pairs of these prime numbers whose sum is a multiple of 5 (such as 5, 10, 15, 20, 25).
3. Check the pairs: $2 + 3 = 5$, $2 + 13 = 15$, and $3 + 7 = 10$, all of which are multiples of 5.

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

> Common mistake: Choosing numbers that are not prime or whose sum does not form a multiple of 5.

### Question 6

*3 marks · Short answer*

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**

1. Examine the list of prime numbers up to 100 to find pairs of primes that are formed by reversing the digits of each other.
2. Identify the given pair 13 and 31.
3. Find the other such pairs up to 100, which are 17 and 71, 37 and 73, and 79 and 97.

**Answer:** The pairs of prime numbers up to 100 with the same digits are 13 and 31, 17 and 71, 37 and 73, and 79 and 97.

> Common mistake: Forgetting to check composite numbers or missing one of the pairs like 79 and 97.

### Question 7

*2 marks · Very short answer*

Find seven consecutive composite numbers between 1 and 100.

**Solution**

1. Seven consecutive composite numbers between 1 and 100 are found by listing numbers that have more than two factors consecutively.
2. These numbers are 90, 91, 92, 93, 94, 95, and 96.

**Answer:** 90, 91, 92, 93, 94, 95, 96

> Common mistake: Listing numbers that are not consecutive or including prime numbers.

### Question 8

*3 marks · Short answer*

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**

1. Twin primes are pairs of prime numbers that have a difference of 2.
2. Checking the prime numbers up to 100, the pairs with a difference of 2 are identified.
3. The other twin primes between 1 and 100 are (3, 5), (5, 7), (11, 13), (17, 19), (29, 31), (41, 43), (59, 61), and (71, 73).

**Answer:** (3, 5), (5, 7), (11, 13), (17, 19), (29, 31), (41, 43), (59, 61), (71, 73)

> Common mistake: Missing some pairs or including pairs that have a difference other than 2.

### Question 9

*3 marks · Short answer*

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.

**Part a**

1. A prime number whose units digit is 4 would be an even number greater than 4.
2. Since 2 is the only even prime number, there is no prime number ending in 4.

Answer a: True

**Part b**

1. Multiplying two or more prime numbers always results in a number with more than two factors.
2. Therefore, a product of primes is always a composite number, never prime.

Answer b: False

**Part c**

1. By definition, a prime number has exactly two factors, which are 1 and the number itself.

Answer c: False

**Part d**

1. While most even numbers are composite, the number 2 is an even prime number.

Answer d: False

**Part e**

1. Every prime number greater than 3 is an odd number.
2. The number immediately following any odd prime is an even number greater than 2, which is always a composite number.

Answer e: True

**Answer:** Statements (a) and (e) are True; statements (b), (c), and (d) are False.

> Common mistake: Forgetting that 2 is an even prime number while evaluating statements about even numbers and primes.

### Question 10

*2 marks · Very short answer*

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

**Solution**

1. Find the prime factorisation of each given number to check the number of distinct prime factors.
2. We get $105 = 3 \times 5 \times 7$, which is the product of exactly three distinct prime numbers.

**Answer:** 105

> Common mistake: Choosing a number like 45 or 60 which has repeated prime factors or fewer distinct factors.

### Question 11

*2 marks · Very short answer*

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

**Solution**

1. The three-digit numbers formed using each of the digits 2, 4, and 5 once are 245, 254, 425, 452, 524, and 542.
2. All these numbers end in 2, 4, or 5, meaning they are divisible by 2 or 5 and thus have more than two factors.
3. Therefore, none of them are prime numbers.

**Answer:** None

> Common mistake: Assuming some combinations can be prime without checking their divisibility by 2 or 5.

### Question 12

*3 marks · Short answer*

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

**Solution**

1. Test prime numbers $p$ such that $2 \times p + 1$ also yields a prime number.
2. For $p = 5$: $2 \times 5 + 1 = 11$ (prime).
3. For $p = 11$: $2 \times 11 + 1 = 23$ (prime).
4. For $p = 23$: $2 \times 23 + 1 = 47$ (prime).
5. For $p = 29$: $2 \times 29 + 1 = 59$ (prime).
6. For $p = 41$: $2 \times 41 + 1 = 83$ (prime).

**Answer:** $11 = 2 \times 5 + 1$, $23 = 2 \times 11 + 1$, $47 = 2 \times 23 + 1$, $59 = 2 \times 29 + 1$, $83 = 2 \times 41 + 1$

> Common mistake: Checking composite numbers instead of prime numbers for $p$.

## Figure it Out

### Question 1

*3 marks · Short answer*

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

**Solution**

1. $64 = 2 \times 2 \times 2 \times 2 \times 2 \times 2$
2. $104 = 2 \times 2 \times 2 \times 13$
3. $105 = 3 \times 5 \times 7$
4. $243 = 3 \times 3 \times 3 \times 3 \times 3$
5. $320 = 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 5$
6. $141 = 3 \times 47$
7. $1728 = 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 3 \times 3 \times 3$
8. $729 = 3 \times 3 \times 3 \times 3 \times 3 \times 3$
9. $1024 = 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2$
10. $1331 = 11 \times 11 \times 11$
11. $1000 = 2 \times 2 \times 2 \times 5 \times 5 \times 5$

**Answer:** Prime factorisations of the given numbers are obtained by repeatedly dividing by prime numbers until only primes are left.

> Common mistake: Stopping division at a composite factor instead of breaking it down into prime factors.

### Question 2

*2 marks · Very short answer*

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

**Solution**

1. Multiply the given prime factors together to find the number.
2. $2 \times 3 \times 3 \times 11 = 198$

**Answer:** 198

> Common mistake: Multiplying the factors incorrectly or missing a factor.

### Question 3

*2 marks · Very short answer*

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

**Solution**

1. Find prime factors of 1955 by checking divisibility by primes less than 30.
2. $1955 = 5 \times 17 \times 23$

**Answer:** $5, 17, 23$

> Common mistake: Trying composite numbers instead of prime numbers.

### Question 4

*3 marks · Short answer*

Find the prime factorisation of these numbers without multiplying first
a. $56 \times 25$
b. $108 \times 75$
c. $1000 \times 81$

**Part a (1 mark)**

1. Write the prime factorisation of 56 and 25 separately.
2. $56 \times 25 = (2 \times 2 \times 2 \times 7) \times (5 \times 5) = 2 \times 2 \times 2 \times 7 \times 5 \times 5$

Answer a: $2 \times 2 \times 2 \times 7 \times 5 \times 5$

**Part b (1 mark)**

1. Write the prime factorisation of 108 and 75 separately.
2. $108 \times 75 = (2 \times 2 \times 3 \times 3 \times 3) \times (3 \times 5 \times 5) = 2 \times 2 \times 3 \times 3 \times 3 \times 3 \times 5 \times 5$

Answer b: $2 \times 2 \times 3 \times 3 \times 3 \times 3 \times 5 \times 5$

**Part c (1 mark)**

1. Write the prime factorisation of 1000 and 81 separately.
2. $1000 \times 81 = (2 \times 2 \times 2 \times 5 \times 5 \times 5) \times (3 \times 3 \times 3 \times 3) = 2 \times 2 \times 2 \times 3 \times 3 \times 3 \times 3 \times 5 \times 5 \times 5$

Answer c: $2 \times 2 \times 2 \times 3 \times 3 \times 3 \times 3 \times 5 \times 5 \times 5$

**Answer:** Prime factorisations found without multiplying first are: (a) $2 \times 2 \times 2 \times 7 \times 5 \times 5$, (b) $2 \times 2 \times 3 \times 3 \times 3 \times 3 \times 5 \times 5$, (c) $2 \times 2 \times 2 \times 3 \times 3 \times 3 \times 3 \times 5 \times 5 \times 5$

> Common mistake: Multiplying the numbers first instead of combining their individual prime factorisations.

### Question 5

*3 marks · Short answer*

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

**Part a (1 mark)**

1. Choose the smallest three different prime numbers: 2, 3, and 5.
2. Multiply them together: $2 \times 3 \times 5 = 30$.

Answer a: 30

**Part b (2 marks)**

1. Choose the smallest four different prime numbers: 2, 3, 5, and 7.
2. Multiply them together: $2 \times 3 \times 5 \times 7 = 210$.

Answer b: 210

**Answer:** The smallest numbers are 30 for three different prime factors and 210 for four different prime factors.

> Common mistake: Using composite numbers instead of prime numbers.

## Figure it Out

### Question 1

*3 marks · Short answer*

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

**Part a (0.75 marks)**

1. Find the prime factorisation of 30: $30 = 2 \times 3 \times 5$
2. Find the prime factorisation of 45: $45 = 3 \times 3 \times 5$
3. Since they have common prime factors 3 and 5, they are not co-prime.

Answer a: No, they are not co-prime.

**Part b (0.75 marks)**

1. Find the prime factorisation of 57: $57 = 3 \times 19$
2. Find the prime factorisation of 85: $85 = 5 \times 17$
3. Since there are no common prime factors, they are co-prime.

Answer b: Yes, they are co-prime.

**Part c (0.75 marks)**

1. Find the prime factorisation of 121: $121 = 11 \times 11$
2. Find the prime factorisation of 1331: $1331 = 11 \times 11 \times 11$
3. Since they have a common prime factor 11, they are not co-prime.

Answer c: No, they are not co-prime.

**Part d (0.75 marks)**

1. Find the prime factorisation of 343: $343 = 7 \times 7 \times 7$
2. Find the prime factorisation of 216: $216 = 2 \times 2 \times 2 \times 3 \times 3 \times 3$
3. Since there are no common prime factors, they are co-prime.

Answer d: Yes, they are co-prime.

**Answer:** Refer to individual parts.

> Common mistake: Conceding that numbers are co-prime by just checking one factor instead of finding the full prime factorisation.

### Question 2

*3 marks · Short answer*

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

**Part a (0.75 marks)**

1. Prime factorisation of 225 is $3 \times 3 \times 5 \times 5$ and of 27 is $3 \times 3 \times 3$.
2. The prime factorisation of 27 is not completely included in the prime factorisation of 225 because 225 contains only two 3s while 27 contains three 3s.
3. Therefore, 225 is not divisible by 27.

Answer a: No

**Part b (0.75 marks)**

1. Prime factorisation of 96 is $2 \times 2 \times 2 \times 2 \times 2 \times 3$ and of 24 is $2 \times 2 \times 2 \times 3$.
2. All prime factors of 24 are included in the prime factorisation of 96.
3. Therefore, 96 is divisible by 24.

Answer b: Yes

**Part c (0.75 marks)**

1. Prime factorisation of 343 is $7 \times 7 \times 7$ and 17 is a prime number.
2. 17 is not a prime factor of 343.
3. Therefore, 343 is not divisible by 17.

Answer c: No

**Part d (0.75 marks)**

1. Prime factorisation of 999 is $3 \times 3 \times 3 \times 37$ and of 99 is $3 \times 3 \times 11$.
2. The prime factors of 99 include 11, which is not a prime factor of 999.
3. Therefore, 999 is not divisible by 99.

Answer d: No

**Answer:** Refer to individual parts.

> Common mistake: Forgetting that a number must include all the prime factors with their correct multiplicities to be divisible.

### Question 3

*3 marks · Short answer*

The first number has prime factorisation $2 \times 3 \times 7$ and the second number has prime factorisation $3 \times 7 \times 11$. Are they co-prime? Does one of them divide the other?

**Solution**

1. Write the given prime factorisations: first number = $2 \times 3 \times 7$, second number = $3 \times 7 \times 11$.
2. Observe that the prime factors 3 and 7 are common to both numbers.
3. Since they share common prime factors, they are not co-prime.
4. Check divisibility by comparing prime factorisations: neither prime factorisation is completely included in the other (the first has 2 and lacks 11, while the second has 11 and lacks 2).
5. Therefore, neither number divides the other.

**Answer:** No, they are not co-prime, and one of them does not divide the other.

> Common mistake: Conclude that they divide each other just because they share some prime factors.

### Question 4

*2 marks · Very short answer*

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

**Solution**

1. Recall that prime numbers have only two factors: 1 and themselves.
2. Since any two distinct prime numbers share no common factors other than 1, they are always co-prime.

**Answer:** Yes, any two prime numbers are co-prime because their only common factor is 1.

> Common mistake: Confusing prime numbers with composite numbers when testing for co-prime pairs.

## Figure it Out

### Question 1

*3 marks · Short answer*

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 born till now, which years were leap years?
b. From the year 2024 till 2099, how many leap years are there?

**Part a (1 mark)**

1. A leap year is a multiple of 4, but century years must be multiples of 400.

Answer a: Answers will vary based on individual birth years.

**Part b (2 marks)**

1. List the leap years from 2024 to 2099, noting that 2100 is not a leap year since it is divisible by 100 but not 400.
2. Count the total number of leap years in this range.
3. The total number of leap years from 2024 till 2099 is 19.

Answer b: 19 leap years

**Answer:** From the year 2024 till 2099, the number of leap years is 19.

> Common mistake: Counting 2100 as a leap year.

### Question 2

*3 marks · Short answer*

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

**Solution**

1. A 4-digit palindrome is of the form $abba$, where $a$ and $b$ are digits ($a \neq 0$).
2. For the number to be divisible by 4, the number formed by its last two digits ($ba$) must be divisible by 4.
3. Testing the digits from largest to smallest, the largest 4-digit palindrome is 8888, and the smallest 4-digit palindrome is 2112.

**Answer:** The largest 4-digit palindrome divisible by 4 is 8888, and the smallest is 2112.

> Common mistake: Choosing 9999 or 1001 without checking divisibility by 4.

### Question 3

*3 marks · Short answer*

Explore and find out if each statement is always true, sometimes true or never true. You can give examples to support your reasoning.
a. Sum of two even numbers gives a multiple of 4.
b. Sum of two odd numbers gives a multiple of 4.

**Part a (1 mark)**

1. Sum of two even numbers is always even, but not always a multiple of 4.
2. Example: $2 + 6 = 8$ (multiple of 4), but $2 + 4 = 6$ (not a multiple of 4).

Answer a: Sometimes true

**Part b (2 marks)**

1. Sum of two odd numbers is always even, but not always a multiple of 4.
2. Example: $1 + 3 = 4$ (multiple of 4), but $1 + 5 = 6$ (not a multiple of 4).

Answer b: Sometimes true

**Answer:** Both statements are sometimes true.

> Common mistake: Assuming the sum of two even numbers is always a multiple of 4.

### Question 4

*3 marks · Short answer*

Find the remainders obtained when each of the following numbers are divided by (a) 10, (b) 5, (c) 2.
78, 99, 173, 572, 980, 1111, 2345

**Solution**

1. For a number, the remainder when divided by 10 is its units digit.
2. The remainder when divided by 5 is the remainder of its units digit divided by 5.
3. The remainder when divided by 2 is 0 if the number is even and 1 if it is odd.

**Answer:** Refer to the remainder table for 78, 99, 173, 572, 980, 1111, and 2345.

> Common mistake: Confusing remainders with the quotient.

### Question 5

*2 marks · Very short answer*

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**

1. A number is divisible by 2, 4, 5, 8, and 10 if it is divisible by their co-prime or highest component prime powers.
2. Checking divisibility by 5 and 8 is sufficient to guarantee divisibility by 2, 4, 5, 8, and 10 because 8 includes 2 and 4, and 5 is co-prime to 8.

**Answer:** 5 and 8

> Common mistake: Choosing two numbers that are not co-prime or do not cover all prime factors.

### Question 6

*2 marks · Very short answer*

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

**Solution**

1. A number is divisible by 2, 4, 5, 8, and 10 if it ends in 0 and the number formed by its last three digits is divisible by 8.
2. Check each given number against these conditions.

**Answer:** 5600, 6000 and 77622160

> Common mistake: Forgetting to check divisibility by 8.

### Question 7

*2 marks · Very short answer*

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

**Solution**

1. Find the prime factorisation of $10000$: $10000 = 2^4 \times 5^4 = 16 \times 625$.
2. The two numbers $16$ and $625$ have units digits $6$ and $5$ respectively, neither of which is $0$.

**Answer:** $16$ and $625$

> Common mistake: Choosing numbers like $100 \times 100$, which have $0$ as their units digit.

## Frequently asked questions

### How many questions are there in NCERT Solutions for Class 6 Maths Chapter 5 Prime Time?

This chapter contains six sets of 'Figure it Out' exercises with a total of 42 questions, consisting of short answer and very short answer types. You can find the complete step-by-step solutions in SwaVid's free PDF available on this page only.

### Which topics are covered in the Class 6 Maths Chapter 5 questions?

The questions cover essential number theory concepts such as factors, multiples, prime and composite numbers, prime factorisation, and common multiples using Venn diagrams. Additional topics include divisibility tests, co-prime numbers, consecutive composite numbers, and even-odd sums.

### What are the hardest question types in Prime Time and how should I approach them?

The most challenging questions involve finding the smallest common multiple of numbers from 1 to 10, determining co-prime numbers using prime factorisation, and working with palindromes or remainders. To approach these, break down the numbers into their prime factors carefully and apply the relevant divisibility rules step by step.

### How do I write answers in Class 6 Maths Chapter 5 to score full marks?

To secure full marks, you should clearly state the mathematical property or rule being used, show the complete prime factorisation, and write down every intermediate step. Referring to SwaVid's expert-verified solutions on this page will help you structure your answers properly.

### Is the free PDF for Class 6 Maths Chapter 5 Prime Time available for download?

Yes, the complete free PDF containing detailed solutions for all exercises in the new NCERT book for the 2026-27 session is available on this page only. You can easily download it to practice offline and prepare thoroughly for your exams.

## Related pages

- [Class 6 Maths chapters](https://www.swavid.com/maths/class/6)

Solutions written by SwaVid, a personal AI tutor for Class 6 to 10 Maths and Science. Practise this chapter free: https://www.swavid.com/start/student