Prime Time — NCERT Solutions, AI Tutor & Practice

Finding factors: Look for pairs that multiply to $36$.

• $1 \times 36$
• $2 \times 18$
• $3 \times 12$
• $4 \times 9$
• $6 \times 6$

So the factors of $36$ are: $1, 2, 3, 4, 6, 9, 12, 18, 36$.

Prime factors among them: A prime has exactly two factors ($1$ and itself).

• $2$ → prime
• $3$ → prime
• ($1$ is not prime; $4, 6, 9, 12, 18, 36$ are composite)

Prime numbers in the list: $2$ and $3$.

A prime number has exactly two factors: $1$ and the number itself.

Check each number and remove those with extra factors:

• $2, 3, 5, 7$ are prime.
• $11, 13, 17, 19$ are prime.
• $23, 29$ are prime.

(Numbers like $4, 6, 8, 9, 10, \dots$ are composite because they have more than two factors.)

Primes between 1 and 30: $2,\ 3,\ 5,\ 7,\ 11,\ 13,\ 17,\ 19,\ 23,\ 29$

There are $10$ such prime numbers.

Use repeated division by the smallest possible prime:

$84 \div 2 = 42$ $42 \div 2 = 21$ $21 \div 3 = 7$ $7 \div 7 = 1$

Collecting the divisors used:

$84 = 2 \times 2 \times 3 \times 7$

Or in exponent form:

$84 = 2^2 \times 3 \times 7$

Two numbers are co-prime if their only common factor is $1$.

Factors of 15: $1, 3, 5, 15$

Factors of 28: $1, 2, 4, 7, 14, 28$

Common factors: only $1$.

Since the only common factor is $1$, 15 and 28 are co-prime.

(Note: co-prime numbers need not themselves be prime — here neither $15$ nor $28$ is prime.)

Twin primes are pairs of prime numbers that differ by $2$.

Primes less than $20$: $2, 3, 5, 7, 11, 13, 17, 19$.

Check which pairs differ by $2$:

• $(3, 5)$ → difference $2$ ✓
• $(5, 7)$ → difference $2$ ✓
• $(11, 13)$ → difference $2$ ✓
• $(17, 19)$ → difference $2$ ✓

Twin prime pairs below 20: $(3,5),\ (5,7),\ (11,13),\ (17,19)$.

Divisibility by 2: last digit is $6$ (even) → divisible by 2 ✓

Divisibility by 3: sum of digits $= 2+3+4+6 = 15$. Since $15$ is divisible by $3$ → divisible by 3 ✓

Divisibility by 6: a number divisible by both $2$ and $3$ is divisible by $6$ → divisible by 6 ✓

Divisibility by 9: digit sum $= 15$, and $15$ is not divisible by $9$ → not divisible by 9 ✗

Conclusion: $2346$ is divisible by $2, 3$ and $6$, but not by $9$.

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

Prime factorisation: $12 = 2^2 \times 3$ $18 = 2 \times 3^2$

Take the highest power of each prime:

• For $2$: highest power is $2^2$
• For $3$: highest power is $3^2$

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

The smallest number divisible by both $12$ and $18$ is $36$.

A composite number has more than two factors.

Numbers between $30$ and $40$: $31, 32, 33, 34, 35, 36, 37, 38, 39$.

Pick three composite ones:

• $32$ → $32 = 2 \times 16$, factors include $1, 2, 4, 8, 16, 32$ (more than two).
• $33$ → $33 = 3 \times 11$, factors are $1, 3, 11, 33$ (more than two).
• $35$ → $35 = 5 \times 7$, factors are $1, 5, 7, 35$ (more than two).

Each has factors other than $1$ and itself, so each is composite.

(Note: $31$ and $37$ are prime and must not be chosen.)

Find the number: $2^3 \times 5 = 8 \times 5 = 40$

Is 40 divisible by 4?

$4 = 2^2$. Since the number contains $2^3$ (which includes $2^2$), it is divisible by $4$.

Check: $40 \div 4 = 10$ ✓

Conclusion: The number is $40$, and it is divisible by 4.

Prime factorisation: $15 = 3 \times 5$ $25 = 5 \times 5 = 5^2$

Common prime factor: $5$ appears in both.

Since $15$ and $25$ share the common factor $5$ (which is greater than $1$), they are not co-prime.

Conclusion: Common prime factor is $5$; the numbers are not co-prime.

Divisible by 2: last digit must be even ($0, 2, 4, 6, 8$).

Divisible by 5: last digit must be $0$ or $5$.

Both conditions together: the last digit must be the value common to both lists → $0$.

So any number divisible by both $2$ and $5$ ends in $0$ (i.e. it is a multiple of $10$).

A number between 40 and 60: $50$ (since $50 \div 2 = 25$ and $50 \div 5 = 10$).

Prime factorisation: $48 = 2^4 \times 3$ $60 = 2^2 \times 3 \times 5$

Take common primes with the lowest power:

• For $2$: lowest power is $2^2$
• For $3$: lowest power is $3^1$
• ($5$ appears only in $60$, so it is not common)

$\text{HCF} = 2^2 \times 3 = 4 \times 3 = 12$

The HCF of $48$ and $60$ is $12$.