Q1: Express each number as a product of its prime factors:

1. 140
   140 ÷ 2 = 70
   70 ÷ 2 = 35
   35 ÷ 5 = 7
   7 is prime.
   Prime factorization: 140 = 2 × 2 × 5 × 7 = 2² × 5 × 7
2. 156
   156 ÷ 2 = 78
   78 ÷ 2 = 39
   39 ÷ 3 = 13
   13 is prime.
   Prime factorization: 156 = 2 × 2 × 3 × 13 = 2² × 3 × 13
3. 3825
   3825 ÷ 3 = 1275
   1275 ÷ 3 = 425
   425 ÷ 5 = 85
   85 ÷ 5 = 17
   17 is prime.
   Prime factorization: 3825 = 3 × 3 × 5 × 5 × 17 = 3² × 5² × 17
4. 5005
   5005 ÷ 5 = 1001
   1001 ÷ 7 = 143
   143 ÷ 11 = 13
   13 is prime.
   Prime factorization: 5005 = 5 × 7 × 11 × 13
5. 7429
   7429 ÷ 7 = 1061
   1061 ÷ 23 = 46.13 (not exact, try next prime)
   1061 ÷ 13 = 81.61 (no)
   1061 ÷ 17 = 62.41 (no)
   1061 ÷ 29 = 36.58 (no)
   1061 is prime.
   So, prime factorization: 7429 = 7 × 1061
   (Note: 1061 is prime)

Q2: Find the LCM and HCF of the following pairs of integers and verify that LCM × HCF = product of the two numbers.

1. 26 and 91
   Prime factors:
   26 = 2 × 13
   91 = 7 × 13
   HCF = 13 (common prime factor)
   LCM = 2 × 7 × 13 = 182
   Check: 182 × 13 = 2366
   Product: 26 × 91 = 2366
   Verified.
2. 510 and 92
   Prime factors:
   510 = 2 × 3 × 5 × 17
   92 = 2 × 2 × 23
   HCF = 2
   LCM = 2 × 2 × 3 × 5 × 17 × 23 = 11790
   Check: 11790 × 2 = 23580
   Product: 510 × 92 = 46920
   Wait, 11790 × 2 = 23580 not equal to 46920, re-check.
   HCF should be product of common prime factors:
   Common prime factors: 2 only
   HCF = 2
   LCM = (510 × 92) / HCF = (510 × 92) / 2 = 23460
   So correct LCM = 23460
   Verification:
   LCM × HCF = 23460 × 2 = 46920 = 510 × 92
   Verified.
3. 336 and 54
   Prime factors:
   336 = 2⁴ × 3 × 7 (since 336 = 16 × 21 = 2⁴ × 3 × 7)
   54 = 2 × 3³ (since 54 = 2 × 27 = 2 × 3³)
   HCF = 2¹ × 3¹ = 6
   LCM = 2⁴ × 3³ × 7 = 3024
   Check: LCM × HCF = 3024 × 6 = 18144
   Product = 336 × 54 = 18144
   Verified.

Q3: Find the LCM and HCF of the following integers by applying the prime factorisation method.

1. 12, 15 and 21
   Prime factors:
   12 = 2² × 3
   15 = 3 × 5
   21 = 3 × 7
   HCF = 3 (common factor)
   LCM = 2² × 3 × 5 × 7 = 420
   
2. 17, 23 and 29
   All are primes, no common factor other than 1.
   HCF = 1
   LCM = 17 × 23 × 29 = 11339
   
3. 8, 9 and 25
   Prime factors:
   8 = 2³
   9 = 3²
   25 = 5²
   HCF = 1
   LCM = 2³ × 3² × 5² = 1800
   

Q4: Given that HCF(306, 657) = 9, find LCM(306, 657).

Using the formula:
LCM × HCF = Product of the two numbers
So,
LCM = (306 × 657) / 9 = 22362

Q5: Check whether 6ⁿ can end with the digit 0 for any natural number n.

Powers of 6 end with the digits:
6¹ = 6
6² = 36
6³ = 216
6⁴ = 1296
6⁵ = 7776
...and so on.
None ends with 0.
To end with 0, the number must be divisible by 10 = 2 × 5.
6ⁿ = 2ⁿ × 3ⁿ, so it has plenty of 2’s but no factor 5.
Hence, 6ⁿ can never end with 0.
Therefore, 6ⁿ never ends with the digit 0.

Q6: Explain why 7 × 11 × 13 + 13 and 7 × 6 × 5 × 4 × 3 × 2 × 1 + 5 are composite numbers.

For the first number:
7 × 11 × 13 + 13 = 13 (7 × 11 + 1) = 13 × (77 + 1) = 13 × 78
Since it is a product of two integers greater than 1, it is composite.

For the second number:
7 × 6 × 5 × 4 × 3 × 2 × 1 + 5 = 7! + 5 = 5040 + 5 = 5045
5045 ÷ 5 = 1009 (since last digit is 5)
Since it is divisible by 5, it is composite.

Q7: There is a circular path around a sports field. Sonia takes 18 minutes to drive one round of the field, while Ravi takes 12 minutes for the same. Suppose they both start at the same point and at the same time, and go in the same direction. After how many minutes will they meet again at the starting point?

They will meet after the LCM of 18 and 12.
Prime factors:
18 = 2 × 3²
12 = 2² × 3
LCM = 2² × 3² = 36
They will meet again after 36 minutes.

Q1: Prove that √5 is irrational.

Proof:

Assume √5 is rational. Then it can be expressed as p/q, where p and q are coprime integers and q ≠ 0.

So,

√5 = p / q

Squaring both sides,

5 = p² / q²

Multiply both sides by q²,

p² = 5 q²

This implies p² is divisible by 5, so p is divisible by 5 (since 5 is prime).

Let p = 5k for some integer k.

Substitute back,

(5k)² = 5 q² → 25 k² = 5 q² → 5 k² = q²

Hence, q² is divisible by 5, so q is divisible by 5.

But this contradicts the assumption that p and q are coprime.

Therefore, √5 is irrational.

Q2: Prove that 3 + 2√5 is irrational.

Proof:

Assume 3 + 2√5 is rational, say equal to r (a rational number).

Then,

3 + 2√5 = r

Rearranging,

2√5 = r - 3

So,

√5 = (r - 3) / 2

The right side is rational (difference and division of rational numbers), but √5 is irrational.

This is a contradiction.

Hence, 3 + 2√5 is irrational.

Q3: Prove the following are irrational:

1. 1 / √2
2. 7√5
3. 6 + √2

Proof:

(i) Suppose 1 / √2 is rational.

Then, √2 = 1 / (1/√2) would be rational (reciprocal of a rational is rational), but √2 is irrational.

Contradiction. So, 1 / √2 is irrational.

(ii) Suppose 7√5 is rational.

Then, √5 = (7√5) / 7 would be rational (division by 7, a rational number), but √5 is irrational.

Contradiction. So, 7√5 is irrational.

(iii) Suppose 6 + √2 is rational, say equal to r.

Then,

6 + √2 = r

Rearranging,

√2 = r - 6

The right side is rational, but √2 is irrational.

Contradiction.

Hence, 6 + √2 is irrational.