NCERT Solutions for Class 6 Maths Chapter 5: Prime Time — Complete Guide

Class 6 Maths Worksheet — Free PDF with Answers

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Why This Chapter Matters: Prime Numbers Are the Atoms of Mathematics

Looking for a class 6 math worksheet or maths worksheet for class 6 with answers? This comprehensive resource covers everything you need. Just as atoms are the building blocks of all matter, prime numbers are the building blocks of all whole numbers. Every number greater than $1$ is either prime or can be broken down into a product of primes in exactly one way. This remarkable fact — called the Fundamental Theorem of Arithmetic — makes prime numbers one of the most important concepts in all of mathematics.

Chapter 5 of the NCERT Class 6 Maths textbook (2024-25 edition) takes you on a journey into the world of primes. You will learn what makes a number prime, how to find all primes up to any limit using the ancient Sieve of Eratosthenes, how to decompose numbers into their prime building blocks, and how to find the HCF and LCM of numbers.

Why should you care about prime numbers?

They simplify fractions. To reduce $\frac{36}{48}$ to its simplest form, you need the HCF of $36$ and $48$. Finding the HCF requires prime factorisation. This is a skill you will use throughout your mathematical life.

They solve real-world problems. Questions like "When will two events that repeat at different intervals happen at the same time?" are LCM problems. Bus schedules, blinking lights, gear rotations — all involve LCM.

They secure the internet. Modern encryption (the technology that keeps your online banking and messaging secure) relies on the mathematical properties of very large prime numbers. The security of the entire internet depends on the fact that multiplying two large primes is easy, but factoring the result back into primes is extremely hard.

They are beautiful. Primes have fascinated mathematicians for over $2{,}000$ years. Questions about primes (like the Goldbach Conjecture and the Riemann Hypothesis) remain unsolved and carry million-dollar prizes.

In this comprehensive guide, we solve every exercise from Chapter 5, explain the underlying concepts in depth, highlight common mistakes, and provide exam strategies and practice problems.

Beyond this worksheet, SparkEd offers interactive online practice for Prime Time with step-by-step solutions and progress tracking — perfect for daily revision.

Key Concepts and Definitions

Let us build a solid foundation with the key concepts of this chapter.

Factors and Multiples

A factor of a number divides it exactly (with no remainder). A multiple of a number is obtained by multiplying it by a whole number.

Example: Factors of $12$: $1, 2, 3, 4, 6, 12$. Multiples of $12$: $12, 24, 36, 48, 60, \ldots$

Key properties:
- $1$ is a factor of every number.
- Every number is a factor of itself.
- Every number is a multiple of itself and of $1$.
- The number of factors is finite. The number of multiples is infinite.
- If $a$ is a factor of $b$, then $b$ is a multiple of $a$.

Finding all factors of $n$: Check each number from $1$ to $\sqrt{n}$. If $d$ divides $n$, then both $d$ and $\frac{n}{d}$ are factors. Stop when $d > \sqrt{n}$.

Example: Factors of $36$:

Factors: $\{1, 2, 3, 4, 6, 9, 12, 18, 36\}$ — nine factors in total.

Prime and Composite Numbers

A prime number is a natural number greater than $1$ that has exactly two factors: $1$ and itself.

Examples: $2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, \ldots$

A composite number is a natural number greater than $1$ that has more than two factors.

Examples: $4, 6, 8, 9, 10, 12, 14, 15, 16, 18, 20, \ldots$

Special cases:
- $1$ is neither prime nor composite. It has exactly one factor (itself).
- $2$ is the only even prime number. Every other even number is divisible by $2$, so it has at least three factors ($1, 2,$ and itself).
- $0$ is neither prime nor composite.

How to check if $n$ is prime: Test divisibility by all primes up to $\sqrt{n}$. If none divide $n$, it is prime.

Example: Is $97$ prime? $\sqrt{97} \approx 9.8$. Check primes up to $9$: $2, 3, 5, 7$.
- $97 \div 2 = 48.5$ (not divisible)
- $97 \div 3 = 32.33...$ (not divisible, digit sum $= 16$, not divisible by $3$)
- $97 \div 5 = 19.4$ (not divisible)
- $97 \div 7 = 13.86...$ (not divisible)

So $97$ is prime.

The Sieve of Eratosthenes

The Sieve of Eratosthenes is an ancient algorithm (over $2{,}200$ years old!) for finding all prime numbers up to a given limit $N$.

Steps:
1. Write all numbers from $2$ to $N$.
2. Circle $2$ (the first prime). Cross out all multiples of $2$ ($4, 6, 8, 10, \ldots$).
3. The next uncrossed number is $3$ (prime). Circle it. Cross out all multiples of $3$ ($9, 15, 21, \ldots$). ($6, 12, 18$ are already crossed.)
4. Next uncrossed: $5$ (prime). Circle it. Cross out multiples of $5$.
5. Continue until you have processed all primes up to $\sqrt{N}$.
6. All remaining uncrossed numbers are prime.

Primes up to $50$: $2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47$ — that is $15$ primes.

Primes up to $100$: There are $25$ primes. The additional ones beyond $50$ are: $53, 59, 61, 67, 71, 73, 79, 83, 89, 97$.

Prime Factorisation

Prime factorisation is the process of expressing a composite number as a product of its prime factors.

Every composite number can be expressed as a product of primes in exactly one way (ignoring the order). This is called the Fundamental Theorem of Arithmetic.

Method 1: Factor Tree
Start with the number. Split it into any two factors. Keep splitting until all factors are prime.

Example: $60$

$$60 = 2^2 \times 3 \times 5$$

Method 2: Division Method
Divide by the smallest possible prime, repeatedly.

$$30 \div 2 = 15$$

$$15 \div 3 = 5$$

$$5 \div 5 = 1$$

So $60 = 2 \times 2 \times 3 \times 5 = 2^2 \times 3 \times 5$.

HCF and LCM

HCF (Highest Common Factor): The largest number that divides two or more numbers exactly. Also called GCD (Greatest Common Divisor).

LCM (Lowest Common Multiple): The smallest positive number that is a multiple of two or more numbers.

Finding HCF using prime factorisation:
1. Find the prime factorisation of each number.
2. Take the lowest power of each common prime factor.
3. Multiply these together.

Finding LCM using prime factorisation:
1. Find the prime factorisation of each number.
2. Take the highest power of each prime factor (whether common or not).
3. Multiply these together.

Key relationship:

This is true for any two positive integers $a$ and $b$.

Coprime numbers: Two numbers are coprime if their HCF is $1$ (they share no common factor other than $1$). Example: $8$ and $15$ are coprime because HCF$(8, 15) = 1$.

Exercise 5.1 — Factors and Multiples

Exercise 5.1 builds your skills in finding factors and multiples of numbers. The systematic factor-pair approach is the key technique.

Solved Example 1: Finding All Factors

Problem: Find all factors of $36$.

Solution:

Check which numbers from $1$ to $\sqrt{36} = 6$ divide $36$:

$$36 = 2 \times 18 \quad \Rightarrow \quad 2 \text{ and } 18$$

$$36 = 3 \times 12 \quad \Rightarrow \quad 3 \text{ and } 12$$

$$36 = 4 \times 9 \quad \Rightarrow \quad 4 \text{ and } 9$$

$$36 = 6 \times 6 \quad \Rightarrow \quad 6$$

We stop at $6$ because $6^2 = 36$.

Factors of $36$: $\{1, 2, 3, 4, 6, 9, 12, 18, 36\}$ — a total of $9$ factors.

Answer: $1, 2, 3, 4, 6, 9, 12, 18, 36$.

Solved Example 2: Finding Common Factors

Problem: Find the common factors of $24$ and $36$.

Solution:

Factors of $24$: $1, 2, 3, 4, 6, 8, 12, 24$
Factors of $36$: $1, 2, 3, 4, 6, 9, 12, 18, 36$

Common factors (numbers appearing in both lists): $1, 2, 3, 4, 6, 12$.

The HCF is the largest common factor $= 12$.

Answer: Common factors are $1, 2, 3, 4, 6, 12$. HCF $= 12$.

Solved Example 3: Multiples and Common Multiples

Problem: Write the first $5$ multiples of $7$ and $9$. Find their LCM.

Solution:

Multiples of $7$: $7, 14, 21, 28, 35, 42, 49, 56, 63, 70, \ldots$
Multiples of $9$: $9, 18, 27, 36, 45, 54, 63, 72, 81, 90, \ldots$

The first common multiple is $63$.

Since $7$ and $9$ are coprime (HCF $= 1$):

Answer: LCM$(7, 9) = 63$.

Solved Example 4: Perfect, Abundant, and Deficient Numbers

Problem: A number is perfect if the sum of its proper factors (all factors except itself) equals the number. Check if $6$ and $28$ are perfect numbers.

Solution:

For $6$: Proper factors: $1, 2, 3$. Sum $= 1 + 2 + 3 = 6$. $\checkmark$ Perfect!

For $28$: Proper factors: $1, 2, 4, 7, 14$. Sum $= 1 + 2 + 4 + 7 + 14 = 28$. $\checkmark$ Perfect!

For comparison, $12$ has proper factors $1, 2, 3, 4, 6$ with sum $16 > 12$ (this is abundant), and $10$ has proper factors $1, 2, 5$ with sum $8 < 10$ (this is deficient).

Answer: Both $6$ and $28$ are perfect numbers.

Solved Example 5: Number of Factors

Problem: Find the number of factors of $72$.

Solution:

Step 1: Prime factorisation.

Step 2: Use the formula: if $n = p_1^{a_1} \times p_2^{a_2} \times \ldots$, then the number of factors is $(a_1 + 1)(a_2 + 1) \ldots$

Number of factors $= (3 + 1)(2 + 1) = 4 \times 3 = 12$.

Step 3: Verify by listing:
$1, 2, 3, 4, 6, 8, 9, 12, 18, 24, 36, 72$ — that is $12$ factors. $\checkmark$

Answer: $72$ has $12$ factors.

Solved Example 6: Factor Pairs

Problem: Write all factor pairs of $48$.

Solution:

$$48 = 2 \times 24$$

$$48 = 3 \times 16$$

$$48 = 4 \times 12$$

$$48 = 6 \times 8$$

Since $\sqrt{48} \approx 6.9$, we stop after $6 \times 8$.

Factor pairs: $(1, 48), (2, 24), (3, 16), (4, 12), (6, 8)$.

Total factors: $10$ (each pair gives $2$ factors).

Answer: Five factor pairs: $(1,48), (2,24), (3,16), (4,12), (6,8)$.

Solved Example 7: Word Problem on Factors

Problem: $48$ students need to be arranged in equal rows. In how many ways can this be done? List all possibilities.

Solution:

The number of rows must be a factor of $48$. Each arrangement corresponds to a factor pair.

| Rows | Students per row |
|------|------------------|
| $1$ | $48$ |
| $2$ | $24$ |
| $3$ | $16$ |
| $4$ | $12$ |
| $6$ | $8$ |
| $8$ | $6$ |
| $12$ | $4$ |
| $16$ | $3$ |
| $24$ | $2$ |
| $48$ | $1$ |

There are $10$ ways.

Answer: $10$ ways to arrange $48$ students in equal rows.

Solved Example 8: Largest Factor Less Than the Number

Problem: What is the largest factor of $100$ that is less than $100$?

Solution:

The largest factor of any number $n$ (other than $n$ itself) is $\frac{n}{p}$ where $p$ is the smallest prime factor of $n$.

Smallest prime factor of $100$ is $2$.

Largest factor less than $100$: $\frac{100}{2} = 50$.

Verify: $100 = 2 \times 50$, so $50$ is indeed a factor. The next largest factor would be $\frac{100}{4} = 25$, which is smaller. $\checkmark$

Answer: $50$.

Exercise 5.2 — Prime and Composite Numbers

Exercise 5.2 focuses on classifying numbers as prime or composite and exploring properties of prime numbers.

Solved Example 1: Identifying Prime Numbers

Problem: Which of the following are prime: $23, 35, 41, 49, 51, 67, 73, 87, 91, 97$?

Solution:

- $23$: Check primes up to $\sqrt{23} \approx 4.8$: $2, 3$. Not divisible. Prime. $\checkmark$
- $35$: $35 = 5 \times 7$. Composite.
- $41$: Check $2, 3, 5$. Not divisible. Prime. $\checkmark$
- $49$: $49 = 7 \times 7$. Composite.
- $51$: $51 = 3 \times 17$. Composite.
- $67$: Check $2, 3, 5, 7$. Not divisible. Prime. $\checkmark$
- $73$: Check $2, 3, 5, 7$. Not divisible. Prime. $\checkmark$
- $87$: $87 = 3 \times 29$. Composite.
- $91$: $91 = 7 \times 13$. Composite. (This one catches many students!)
- $97$: Check $2, 3, 5, 7$. Not divisible. Prime. $\checkmark$

Answer: $23, 41, 67, 73, 97$ are prime.

Solved Example 2: The Sieve of Eratosthenes up to 50

Problem: Use the Sieve of Eratosthenes to find all primes up to $50$.

Solution:

Step 1: Write numbers $2$ to $50$.

Step 2: $\sqrt{50} \approx 7.07$. So we sieve using primes $2, 3, 5, 7$.

Cross out multiples of $2$ (keep $2$): $4, 6, 8, 10, 12, 14, 16, 18, 20, 22, 24, 26, 28, 30, 32, 34, 36, 38, 40, 42, 44, 46, 48, 50$.

Cross out multiples of $3$ (keep $3$): $9, 15, 21, 27, 33, 39, 45$ (others already crossed).

Cross out multiples of $5$ (keep $5$): $25, 35$ (others already crossed).

Cross out multiples of $7$ (keep $7$): $49$ (others already crossed).

Primes up to $50$: $2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47$.

That is $15$ primes.

Answer: $15$ primes up to $50$: $2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47$.

Solved Example 3: Twin Primes

Problem: Twin primes are pairs of primes that differ by $2$. Find all twin prime pairs up to $50$.

Solution:

From our list of primes up to $50$: $2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47$.

Check consecutive primes for a difference of $2$:
- $(3, 5)$: $5 - 3 = 2$ $\checkmark$
- $(5, 7)$: $7 - 5 = 2$ $\checkmark$
- $(11, 13)$: $13 - 11 = 2$ $\checkmark$
- $(17, 19)$: $19 - 17 = 2$ $\checkmark$
- $(29, 31)$: $31 - 29 = 2$ $\checkmark$
- $(41, 43)$: $43 - 41 = 2$ $\checkmark$

Answer: Twin prime pairs up to $50$: $(3,5), (5,7), (11,13), (17,19), (29,31), (41,43)$.

Solved Example 4: Goldbach's Conjecture

Problem: Goldbach's Conjecture states that every even number greater than $2$ can be written as the sum of two primes. Verify this for $4, 6, 8, 10, 20, 30$.

Solution:

$$6 = 3 + 3$$

$$8 = 3 + 5$$

$$10 = 3 + 7 = 5 + 5$$

$$20 = 3 + 17 = 7 + 13$$

$$30 = 7 + 23 = 11 + 19 = 13 + 17$$

All verified! $\checkmark$

Fun fact: This conjecture has been verified for all even numbers up to $4 \times 10^{18}$ (that is $4$ followed by $18$ zeros!) but has never been mathematically proven. It remains one of the oldest unsolved problems in mathematics.

Answer: Verified for all given numbers.

Solved Example 5: Consecutive Primes

Problem: Can three consecutive natural numbers all be prime? What about two?

Solution:

Three consecutive numbers: Among any three consecutive numbers, one must be divisible by $3$. The only multiple of $3$ that is prime is $3$ itself. So the only possibility is $1, 2, 3$ or $2, 3, 4$.

$1$ is not prime. $4$ is not prime. So $2, 3$ is the only pair of consecutive primes that are also part of a triple.

However, $(3, 4, 5)$ has $4$ as composite. No triple of consecutive numbers (all $\geq 2$) can be all prime.

Wait: $(2, 3)$ are consecutive primes. But there are no three consecutive primes.

Actually: $2, 3, 5$ are primes but not consecutive numbers. $3, 5, 7$ are primes but not consecutive numbers (they differ by $2$).

Three consecutive natural numbers that are ALL prime: impossible (since one of every three consecutive numbers is divisible by $3$, and if that number is $> 3$, it is composite).

Two consecutive primes that are consecutive numbers: Only $(2, 3)$. For any $n > 2$, one of $n, n+1$ must be even (hence composite if $> 2$).

Answer: The only consecutive natural numbers that are both prime are $2$ and $3$. No three consecutive numbers can all be prime.

Solved Example 6: Primes Between Two Numbers

Problem: How many prime numbers are there between $30$ and $50$?

Solution:

Check each odd number between $30$ and $50$ (even numbers $> 2$ are composite):

- $31$: not divisible by $2, 3, 5$. $\sqrt{31} \approx 5.6$. Prime. $\checkmark$
- $33 = 3 \times 11$. Composite.
- $35 = 5 \times 7$. Composite.
- $37$: not divisible by $2, 3, 5$. Prime. $\checkmark$
- $39 = 3 \times 13$. Composite.
- $41$: not divisible by $2, 3, 5$. Prime. $\checkmark$
- $43$: not divisible by $2, 3, 5$. Prime. $\checkmark$
- $45 = 5 \times 9$. Composite.
- $47$: not divisible by $2, 3, 5$. Prime. $\checkmark$
- $49 = 7 \times 7$. Composite.

Answer: There are $5$ primes between $30$ and $50$: $31, 37, 41, 43, 47$.

Exercise 5.3 — Prime Factorisation

Exercise 5.3 teaches you to express numbers as products of prime factors using the factor tree and division methods. This is the most important skill in the chapter — it is used to find HCF, LCM, simplify fractions, and much more.

Solved Example 1: Factor Tree Method

Problem: Find the prime factorisation of $180$ using a factor tree.

Solution:

$$90 = 2 \times 45$$

$$45 = 3 \times 15$$

$$15 = 3 \times 5$$

Collecting all the prime factors:

Verification: $4 \times 9 \times 5 = 180$ $\checkmark$

Answer: $180 = 2^2 \times 3^2 \times 5$.

Solved Example 2: Division Method

Problem: Find the prime factorisation of $252$ using the division method.

Solution:

$$126 \div 2 = 63$$

$$63 \div 3 = 21$$

$$21 \div 3 = 7$$

$$7 \div 7 = 1$$

Reading the divisors: $252 = 2 \times 2 \times 3 \times 3 \times 7 = 2^2 \times 3^2 \times 7$.

Verification: $4 \times 9 \times 7 = 252$ $\checkmark$

Answer: $252 = 2^2 \times 3^2 \times 7$.

Solved Example 3: Comparing Two Methods

Problem: Find the prime factorisation of $120$ using both methods.

Solution:

Factor tree:

Division method:

Both give: $120 = 2^3 \times 3 \times 5$.

This confirms the Fundamental Theorem of Arithmetic: the prime factorisation is unique regardless of the method used.

Answer: $120 = 2^3 \times 3 \times 5$.

Solved Example 4: Large Number Factorisation

Problem: Find the prime factorisation of $1{,}260$.

Solution:

$$630 \div 2 = 315$$

$$315 \div 3 = 105$$

$$105 \div 3 = 35$$

$$35 \div 5 = 7$$

$$7 \div 7 = 1$$

Verification: $4 \times 9 \times 5 \times 7 = 4 \times 9 \times 35 = 4 \times 315 = 1{,}260$ $\checkmark$

Answer: $1{,}260 = 2^2 \times 3^2 \times 5 \times 7$.

Solved Example 5: Index Form

Problem: Express $3{,}600$ in prime factorisation (index form).

Solution:

$$225 \div 3 = 75, \quad 75 \div 3 = 25, \quad 25 \div 5 = 5, \quad 5 \div 5 = 1$$

Verification: $16 \times 9 \times 25 = 16 \times 225 = 3{,}600$ $\checkmark$

Number of factors: $(4+1)(2+1)(2+1) = 5 \times 3 \times 3 = 45$ factors.

Answer: $3{,}600 = 2^4 \times 3^2 \times 5^2$.

Exercise 5.4 — HCF and LCM

Exercise 5.4 brings together everything you have learned to find the HCF and LCM of numbers. These are among the most frequently tested concepts in exams.

Solved Example 1: HCF Using Prime Factorisation

Problem: Find the HCF of $36$ and $48$.

Solution:

Prime factorisations:

$$48 = 2^4 \times 3$$

HCF = product of common primes with lowest powers:
- Common prime $2$: lowest power is $\min(2, 4) = 2 \Rightarrow 2^2$
- Common prime $3$: lowest power is $\min(2, 1) = 1 \Rightarrow 3^1$

Verification: $36 \div 12 = 3$ $\checkmark$ and $48 \div 12 = 4$ $\checkmark$.

Answer: HCF$(36, 48) = 12$.

Solved Example 2: LCM Using Prime Factorisation

Problem: Find the LCM of $36$ and $48$.

Solution:

Prime factorisations (from above):

$$48 = 2^4 \times 3$$

LCM = product of all primes with highest powers:
- Prime $2$: highest power is $\max(2, 4) = 4 \Rightarrow 2^4$
- Prime $3$: highest power is $\max(2, 1) = 2 \Rightarrow 3^2$

Verification using the HCF-LCM relationship:

$$a \times b = 36 \times 48 = 1{,}728$$

$\checkmark$

Answer: LCM$(36, 48) = 144$.

Solved Example 3: HCF and LCM of Three Numbers

Problem: Find the HCF and LCM of $12$, $18$, and $24$.

Solution:

Prime factorisations:

$$18 = 2 \times 3^2$$

$$24 = 2^3 \times 3$$

HCF: Take lowest powers of common primes:
- $2$: $\min(2, 1, 3) = 1$
- $3$: $\min(1, 2, 1) = 1$

LCM: Take highest powers of all primes:
- $2$: $\max(2, 1, 3) = 3$
- $3$: $\max(1, 2, 1) = 2$

Answer: HCF $= 6$, LCM $= 72$.

Solved Example 4: Word Problem — LCM Application

Problem: Two bells ring at intervals of $12$ minutes and $18$ minutes. If they ring together at $12{:}00$ PM, when will they next ring together?

Solution:

We need the LCM of $12$ and $18$.

$$\text{LCM} = 2^2 \times 3^2 = 4 \times 9 = 36$$

The bells will next ring together after $36$ minutes.

$12{:}00$ PM $+ 36$ minutes $= 12{:}36$ PM.

Answer: They next ring together at $12{:}36$ PM.

Solved Example 5: Word Problem — HCF Application

Problem: A rectangular room measures $12$ m by $18$ m. What is the largest square tile that can cover the floor without cutting?

Solution:

The side of the largest square tile must divide both $12$ and $18$ exactly. This is the HCF.

The largest tile has side $6$ m.

Number of tiles needed: $\frac{12 \times 18}{6 \times 6} = \frac{216}{36} = 6$ tiles.

Answer: Largest square tile has side $6$ m. Six tiles are needed.

Solved Example 6: Simplifying Fractions Using HCF

Problem: Reduce $\frac{84}{126}$ to its simplest form.

Solution:

Find HCF$(84, 126)$:

$$126 = 2 \times 3^2 \times 7$$

$$\text{HCF} = 2 \times 3 \times 7 = 42$$

Divide numerator and denominator by $42$:

Answer: $\frac{84}{126} = \frac{2}{3}$.

Solved Example 7: Finding a Number from HCF and LCM

Problem: The HCF of two numbers is $6$ and their LCM is $180$. If one number is $36$, find the other.

Solution:

Using the relationship: $\text{HCF} \times \text{LCM} = a \times b$.

$$1080 = 36b$$

$$b = \frac{1080}{36} = 30$$

Verify: HCF$(36, 30)$:
$36 = 2^2 \times 3^2$, $30 = 2 \times 3 \times 5$.
HCF $= 2 \times 3 = 6$ $\checkmark$
LCM $= 2^2 \times 3^2 \times 5 = 180$ $\checkmark$

Answer: The other number is $30$.

Solved Example 8: Coprime Numbers

Problem: Are the following pairs coprime? (a) $15$ and $28$ (b) $12$ and $35$ (c) $14$ and $21$

Solution:

Two numbers are coprime if their HCF is $1$.

(a) $15 = 3 \times 5$, $28 = 2^2 \times 7$. No common prime factor. HCF $= 1$. Coprime. $\checkmark$

(b) $12 = 2^2 \times 3$, $35 = 5 \times 7$. No common prime factor. HCF $= 1$. Coprime. $\checkmark$

(c) $14 = 2 \times 7$, $21 = 3 \times 7$. Common factor: $7$. HCF $= 7 \neq 1$. Not coprime.

Answer: (a) and (b) are coprime; (c) is not coprime.

Common Mistakes and How to Avoid Them

Here are the most frequent errors in this chapter:

1. Treating $1$ as Prime
* Mistake: Listing $1$ as a prime number.
* Fix: $1$ is NEITHER prime NOR composite. A prime number must have exactly two distinct factors. $1$ has only one factor (itself).

2. Forgetting $2$ Is Prime
* Mistake: Saying "all primes are odd" and skipping $2$.
* Fix: $2$ is the smallest prime and the ONLY even prime. Always include it.

3. Stopping Prime Factorisation Too Early
* Mistake: Writing $36 = 4 \times 9$ and stopping (since $4$ and $9$ are not prime).
* Fix: Continue until ALL factors are prime: $36 = 2^2 \times 3^2$.

4. Confusing HCF and LCM
* Mistake: Taking highest powers for HCF or lowest powers for LCM.
* Fix: HCF = Highest Common Factor uses lowest powers of common primes. LCM = Lowest Common Multiple uses highest powers of all primes.

5. Thinking $91$ Is Prime
* Mistake: Assuming $91$ is prime because it looks prime.
* Fix: Always check! $91 = 7 \times 13$. Test divisibility by primes up to $\sqrt{91} \approx 9.5$: check $2, 3, 5, 7$.

6. Using the Wrong Relationship
* Mistake: Writing HCF $+$ LCM $= a \times b$.
* Fix: The correct relationship is HCF $\times$ LCM $= a \times b$.

7. Not Verifying the Answer
* Mistake: Finding HCF or LCM without checking.
* Fix: Verify: the HCF must divide both numbers. The LCM must be a multiple of both numbers.

Exam Strategy: Scoring Full Marks in Prime Time

Practice Problems — Try These Yourself

Problem 1: Find all factors of $60$. How many factors does it have?

Problem 2: Is $119$ prime or composite? Justify your answer.

Problem 3: Find the prime factorisation of $540$ using the division method.

Problem 4: Find the HCF and LCM of $42$ and $70$.

Problem 5: Three lights blink every $6$, $8$, and $12$ seconds. If they all blink together at time $0$, after how many seconds will they next blink together?

Problem 6: A rope of $36$ m and another of $48$ m are to be cut into pieces of equal length. What is the greatest possible length of each piece?

Problem 7: Simplify $\frac{72}{108}$ using HCF.

Problem 8: The HCF of two numbers is $8$ and their LCM is $240$. If one number is $48$, find the other.

Problem 9: Find all twin prime pairs between $50$ and $100$.

Problem 10: Using the Sieve of Eratosthenes, find how many primes are between $1$ and $100$.

Problem 11: Are $35$ and $72$ coprime? Justify.

Problem 12: Express $2{,}520$ as a product of prime factors in index form.

Quick Revision Notes

Factors and Multiples:
- Factor divides exactly (finite count)
- Multiple is obtained by multiplying (infinite count)
- $1$ is a factor of every number
- Find factors by checking up to $\sqrt{n}$

Prime and Composite:
- Prime: exactly $2$ factors ($1$ and itself)
- Composite: more than $2$ factors
- $1$: neither prime nor composite
- $2$: only even prime

Primes up to $50$: $2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47$ ($15$ primes)

Prime Factorisation:
- Every composite number = unique product of primes
- Use division method or factor tree
- Write in index form: $360 = 2^3 \times 3^2 \times 5$

HCF (Highest Common Factor):
- Lowest powers of COMMON primes
- Use for: largest piece, greatest length, simplifying fractions

LCM (Lowest Common Multiple):
- Highest powers of ALL primes
- Use for: simultaneous events, "together again" problems

Key Relationship:

Number of factors: If $n = p_1^{a_1} \times p_2^{a_2} \times \ldots$, number of factors $= (a_1+1)(a_2+1)\ldots$

Coprime: HCF $= 1$ (no common prime factor)

Boost Your Preparation with SparkEd

You have now worked through every exercise in Chapter 5 — from finding factors to the Sieve of Eratosthenes to computing HCF and LCM. These are skills you will use in almost every math chapter going forward.

Here is how SparkEd can help:

- Practice by Difficulty: On our Prime Time practice page, work through Level 1, 2, and 3 problems.

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- AI Coach: Get personalised recommendations on which concepts need more practice.

- Cross-Topic Connections: Prime Time connects to Number Play (divisibility rules) and Fractions (simplifying fractions using HCF). Explore all chapters on our programs page.

Head over to sparkedmaths.com and start practising today!