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

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 $36$:

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 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 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: $60$

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

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

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$(8, 15) = 1$.

Problem: Find all factors of $36$.

Solution:

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

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

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 $= 12$.

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$(7, 9) = 63$.

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.

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.

Problem: Write all factor pairs of $48$.

Solution:

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

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.

There are $10$ ways.

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

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: $50$.

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.

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

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 $50$: $(3,5), (5,7), (11,13), (17,19), (29,31), (41,43)$.

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:

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.

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.

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 $50$: $31, 37, 41, 43, 47$.

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

Solution:

Collecting all the prime factors:

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

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

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

Solution:

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

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

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

Solution:

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

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

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

Solution:

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

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

Solution:

Prime factorisations:

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$(36, 48) = 12$.

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

Solution:

Prime factorisations (from above):

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:

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

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

Solution:

Prime factorisations:

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 $= 72$.

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 $18$.

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.

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.

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

Solution:

Find HCF$(84, 126)$:

Divide numerator and denominator by $42$:

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

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

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 $30$.

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.