Solution:

Let a be any positive integer, b = 3. Then a = 3q + r, where r = 0, 1, 2.

Case 1: a = 3q → a² = 9q² = 3(3q²) = 3m

Case 2: a = 3q + 1 → a² = (3q+1)² = 9q² + 6q + 1 = 3(3q²+2q) + 1 = 3m + 1

Case 3: a = 3q + 2 → a² = (3q+2)² = 9q² + 12q + 4 = 3(3q²+4q+1) + 1 = 3m + 1

∴ Square of any positive integer is of the form 3m or 3m+1.

Solution:

Let a be any positive integer, b = 4. Then a = 4q + r, where r = 0, 1, 2, 3.

a = 4q (even), 4q + 1 (odd), 4q + 2 (even), 4q + 3 (odd)

∴ Odd integers are of the form 4q + 1 or 4q + 3.

Solution:

Let a be any positive integer, b = 6. Then a = 6q + r, where r = 0, 1, 2, 3, 4, 5.

6q (even), 6q+1 (odd), 6q+2 (even), 6q+3 (odd), 6q+4 (even), 6q+5 (odd)

∴ Odd integers of form 6q+1, 6q+3, 6q+5.

Solution:

(i) 135 and 225:
225 = 135 × 1 + 90
135 = 90 × 1 + 45
90 = 45 × 2 + 0
∴ HCF = 45

(ii) 196 and 38220:
38220 = 196 × 195 + 0
∴ HCF = 196

(iii) 867 and 255:
867 = 255 × 3 + 102
255 = 102 × 2 + 51
102 = 51 × 2 + 0
∴ HCF = 51

Solution:

Let a be any positive integer, b = 6. Then a = 6q + r, 0 ≤ r < 6, r = 0,1,2,3,4,5

6q (even), 6q+1(odd), 6q+2(even), 6q+3(odd), 6q+4(even), 6q+5(odd)

∴ Odd integers = 6q+1, 6q+3, 6q+5.

Solution:

Max columns = HCF of 616 and 32

616 = 32 × 19 + 8

32 = 8 × 4 + 0

∴ HCF = 8

Maximum columns = 8

Solution:

Let a be any +ve integer, b = 3. Then a = 3q + r, r = 0,1,2

a=3q → a²=9q²=3(3q²)=3m

a=3q+1 → a²=9q²+6q+1=3(3q²+2q)+1=3m+1

a=3q+2 → a²=9q²+12q+4=3(3q²+4q+1)+1=3m+1

∴ Square is of form 3m or 3m+1.

Solution:

Let a be any +ve integer, b = 3. Then a = 3q + r, r = 0,1,2

Case r=0: a=3q, a³=27q³=9(3q³)=9m

Case r=1: a=3q+1, a³=27q³+27q²+9q+1=9(3q³+3q²+q)+1=9m+1

Case r=2: a=3q+2, a³=27q³+54q²+36q+8=9(3q³+6q²+4q)+8=9m+8

∴ Cube is of form 9m, 9m+1 or 9m+8.

Solution:

(i) 140 = 2 × 2 × 5 × 7 = 2² × 5 × 7

(ii) 156 = 2 × 2 × 3 × 13 = 2² × 3 × 13

(iii) 3825 = 3 × 3 × 5 × 5 × 17 = 3² × 5² × 17

(iv) 5005 = 5 × 7 × 11 × 13

(v) 7429 = 17 × 19 × 23

Solution:

(i) 26 and 91:
26 = 2 × 13, 91 = 7 × 13
HCF = 13, LCM = 2 × 7 × 13 = 182
Check: 13 × 182 = 2366, 26 × 91 = 2366 ✓

(ii) 510 and 92:
510 = 2 × 3 × 5 × 17, 92 = 2² × 23
HCF = 2, LCM = 2² × 3 × 5 × 17 × 23 = 23460
Check: 2 × 23460 = 46920, 510 × 92 = 46920 ✓

(iii) 336 and 54:
336 = 2⁴ × 3 × 7, 54 = 2 × 3³
HCF = 2 × 3 = 6, LCM = 2⁴ × 3³ × 7 = 3024
Check: 6 × 3024 = 18144, 336 × 54 = 18144 ✓

Solution:

(i) 12, 15, 21:
12 = 2² × 3, 15 = 3 × 5, 21 = 3 × 7
HCF = 3, LCM = 2² × 3 × 5 × 7 = 420

(ii) 17, 23, 29 (all prime):
HCF = 1, LCM = 17 × 23 × 29 = 11339

(iii) 8, 9, 25:
8 = 2³, 9 = 3², 25 = 5²
HCF = 1, LCM = 2³ × 3² × 5² = 1800

Solution:

LCM × HCF = a × b

LCM × 9 = 306 × 657

LCM = (306 × 657)/9 = (306/9) × 657 = 34 × 657 = 22338

Solution:

For 6ⁿ to end with 0, it must be divisible by 10 = 2 × 5.

6ⁿ = (2 × 3)ⁿ = 2ⁿ × 3ⁿ

It has only 2 and 3 as prime factors, not 5.

∴ 6ⁿ cannot end with digit 0 for any n.

Solution:

First number: 7×11×13 + 13 = 13(7×11 + 1) = 13 × 78 → composite

Second number: 7×6×5×4×3×2×1 + 5 = 5(7×6×4×3×2×1 + 1) = 5 × 1009 → composite

Solution:

They meet after LCM of 18 and 12 minutes.

18 = 2 × 3², 12 = 2² × 3

LCM = 2² × 3² = 4 × 9 = 36

They meet after 36 minutes.

Solution:

Let √5 = a/b, where a,b are co-prime integers, b≠0.

Squaring: 5 = a²/b² → a² = 5b²

∴ a is divisible by 5. Let a = 5c.

Then (5c)² = 5b² → 25c² = 5b² → b² = 5c²

∴ b is also divisible by 5.

This contradicts a and b being co-prime.

∴ √5 is irrational.

Solution:

Let 3 + 2√5 = a/b, rational.

2√5 = a/b - 3 = (a - 3b)/b

√5 = (a - 3b)/2b (rational)

This contradicts √5 being irrational.

∴ 3 + 2√5 is irrational.

Solution:

Let √2 = a/b, a,b co-prime, b≠0.

Squaring: 2 = a²/b² → a² = 2b²

∴ a is even. Let a = 2c.

(2c)² = 2b² → 4c² = 2b² → b² = 2c²

∴ b is also even.

This contradicts a,b being co-prime.

∴ √2 is irrational.

Solution:

Terminating: Denominator of form 2ᵐ×5ⁿ

(i) 13/3125 → 3125=5⁵ → Terminating

(ii) 17/8 → 8=2³ → Terminating

(iii) 64/455 → 455=5×7×13 → Non-terminating

(iv) 15/1600 → 1600=2⁶×5² → Terminating

(v) 29/343 → 343=7³ → Non-terminating

(vi) 23/(2³×5²) → Terminating

(vii) 129/(2²×5⁷×7⁵) → Non-terminating

(viii) 6/15 = 2/5 → Terminating

(ix) 35/50 = 7/10 → Terminating

(x) 77/210 = 11/30 → 30=2×3×5 → Non-terminating

Solution:

(i) 13/3125 = 13 × 2⁵ / 3125 × 2⁵ = 416/100000 = 0.00416

(ii) 17/8 = 17 × 125 / 8 × 125 = 2125/1000 = 2.125

(iii) 64/455 → Non-terminating = 0.140659...

(iv) 15/1600 = 15/16×100 = 0.9375/100 = 0.009375

(v) 29/343 → Non-terminating = 0.084548...

Solution:

If x = p/q has terminating decimal expansion, then q must be of the form 2ᵐ × 5ⁿ.

455 = 42 × 10 + 35

42 = 35 × 1 + 7

35 = 7 × 5 + 0

∴ HCF = 7

210 = 55 × 3 + 45

55 = 45 × 1 + 10

45 = 10 × 4 + 5

10 = 5 × 2 + 0

∴ HCF = 5

120 = 2³ × 3 × 5

144 = 2⁴ × 3²

LCM = 2⁴ × 3² × 5 = 16 × 9 × 5 = 720

6 = 2 × 3

72 = 2³ × 3²

120 = 2³ × 3 × 5

HCF = 2 × 3 = 6

LCM = 2³ × 3² × 5 = 8 × 9 × 5 = 360

Let √3 = a/b (co-prime)

3 = a²/b² → a² = 3b²

∴ a is divisible by 3. Let a = 3c.

(3c)² = 3b² → 9c² = 3b² → b² = 3c²

∴ b is also divisible by 3. Contradiction.

∴ √3 is irrational.

156 = 126 × 1 + 30

126 = 30 × 4 + 6

30 = 6 × 5 + 0

∴ HCF = 6

Same method as √5. Let √7 = a/b, a,b co-prime.

7 = a²/b² → a² = 7b² → a divisible by 7 → b divisible by 7 → Contradiction.

∴ √7 is irrational.

225 = 135 × 1 + 90

135 = 90 × 1 + 45

90 = 45 × 2 + 0

∴ HCF = 45

3825 = 3 × 3 × 5 × 5 × 17 = 3² × 5² × 17

24 = 2³ × 3, 36 = 2² × 3²

HCF = 2² × 3 = 12, LCM = 2³ × 3² = 72

Check: 12 × 72 = 864, 24 × 36 = 864 ✓