Solution:

Euclid’s Division Lemma states that for integers a and b, there exist unique integers q and r such that a = bq + r, where 0 ≤ r < b.

• Step 1: Divide 168 by 96: 168 = 96 × 1 + 72.
• Step 2: Divide 96 by 72: 96 = 72 × 1 + 24.
• Step 3: Divide 72 by 24: 72 = 24 × 3 + 0.
• HCF is the last non-zero remainder: 24.

Answer: HCF of 96 and 168 is 24.

Solution:

By Euclid’s Division Lemma, any integer n divided by 2 gives quotient q and remainder r, where r = 0 or 1.

• Case 1: If r = 0, then n = 2q (even, e.g., 4 = 2 × 2).
• Case 2: If r = 1, then n = 2q + 1 (odd, e.g., 5 = 2 × 2 + 1).
• Since r can only be 0 or 1, all positive integers are either 2q or 2q + 1.

Answer: Any positive integer is of the form 2q or 2q + 1.

Solution:

The maximum number of items per box is the HCF of 150 and 225.

• Find HCF: 225 = 150 × 1 + 75.
• 150 = 75 × 2 + 0.
• HCF = 75.
• Number of boxes: Pens = 150 ÷ 75 = 2, Pencils = 225 ÷ 75 = 3.

Answer: Maximum number of items per box is 75.

Solution:

Any integer n divided by 3 has remainder r = 0, 1, or 2, so n = 3q, 3q + 1, or 3q + 2.

• Case 1: n = 3q. Square: n² = (3q)² = 9q² = 3(3q²) = 3m, where m = 3q².
• Case 2: n = 3q + 1. Square: n² = (3q + 1)² = 9q² + 6q + 1 = 3(3q² + 2q) + 1 = 3m + 1, where m = 3q² + 2q.
• Case 3: n = 3q + 2. Square: n² = (3q + 2)² = 9q² + 12q + 4 = 9q² + 12q + 3 + 1 = 3(3q² + 4q + 1) + 1 = 3m + 1, where m = 3q² + 4q + 1.
• Thus, n² is either 3m or 3m + 1.

Answer: The square of any positive integer is of the form 3m or 3m + 1.

Solution:

Any integer n divided by 3 has remainder r = 0, 1, or 2, so n = 3q, 3q + 1, or 3q + 2.

• Case 1: n = 3q. Cube: n³ = (3q)³ = 27q³ = 9(3q³) = 9m, where m = 3q³.
• Case 2: n = 3q + 1. Cube: n³ = (3q + 1)³ = 27q³ + 27q² + 9q + 1 = 9(3q³ + 3q² + q) + 1 = 9m + 1, where m = 3q³ + 3q² + q.
• Case 3: n = 3q + 2. Cube: n³ = (3q + 2)³ = 27q³ + 54q² + 36q + 8 = 9(3q³ + 6q² + 4q) + 8 = 9m + 8, where m = 3q³ + 6q² + 4q.
• Thus, n³ is either 9m, 9m + 1, or 9m + 8.

Answer: The cube of any positive integer is of the form 9m, 9m + 1, or 9m + 8.

Solution:

The Fundamental Theorem of Arithmetic states that every composite number has a unique prime factorization.

• Divide 360 by 2: 360 ÷ 2 = 180.
• Divide 180 by 2: 180 ÷ 2 = 90.
• Divide 90 by 2: 90 ÷ 2 = 45.
• Divide 45 by 3: 45 ÷ 3 = 15.
• Divide 15 by 3: 15 ÷ 3 = 5.
• 5 is prime.
• Thus, 360 = 2 × 2 × 2 × 3 × 3 × 5 = 2³ × 3² × 5.

Answer: 360 = 2³ × 3² × 5.

Solution:

Use prime factorization to find LCM and HCF.

• Prime factorization: 12 = 2² × 3, 28 = 2² × 7.
• HCF = 2² = 4 (lowest powers).
• LCM = 2² × 3 × 7 = 4 × 3 × 7 = 84 (highest powers).

Answer: HCF = 4, LCM = 84.

Solution:

Use prime factorization to find LCM and HCF, then verify for one pair (e.g., 16 and 24).

• Prime factorization: 16 = 2⁴, 24 = 2³ × 3, 40 = 2³ × 5.
• HCF = 2³ = 8 (lowest powers).
• LCM = 2⁴ × 3 × 5 = 16 × 3 × 5 = 240 (highest powers).
• Verify for 16 and 24: LCM = 2⁴ × 3 = 48, HCF = 2³ = 8.
• LCM × HCF = 48 × 8 = 384, Product = 16 × 24 = 384.
• Since 48 × 8 = 16 × 24, the relationship holds.

Answer: HCF = 8, LCM = 240. LCM × HCF = 16 × 24 for the pair (16, 24).

Solution:

Use Euclid’s Division Lemma to find HCF, then express it as a linear combination.

• Step 1: 180 = 108 × 1 + 72.
• Step 2: 108 = 72 × 1 + 36.
• Step 3: 72 = 36 × 2 + 0.
• HCF = 36.
• Linear combination: From 108 = 72 × 1 + 36, we get 36 = 108 – 72.
• From 180 = 108 × 1 + 72, we get 72 = 180 – 108.
• Substitute: 36 = 108 – (180 – 108) = 108 – 180 + 108 = 2 × 108 – 180.
• Thus, 36 = 108 × 2 + 180 × (–1).

Answer: HCF = 36, expressed as 36 = 108 × 2 + 180 × (–1).

Solution:

By Euclid’s Division Lemma, any integer n divided by 6 has remainder r = 0, 1, 2, 3, 4, or 5, so n = 6q, 6q + 1, …, 6q + 5.

• Check for odd numbers:
• 6q: Even (divisible by 2).
• 6q + 1: Odd (e.g., 1, 7).
• 6q + 2: Even (divisible by 2).
• 6q + 3: Odd (e.g., 3, 9).
• 6q + 4: Even (divisible by 2).
• 6q + 5: Odd (e.g., 5, 11).
• Thus, odd integers are of the form 6q + 1, 6q + 3, or 6q + 5.

Answer: Any positive odd integer is of the form 6q + 1, 6q + 3, or 6q + 5.

Solution:

Let the two consecutive integers be n and n + 1.

• One of n or n + 1 must be even (since consecutive integers alternate between odd and even).
• If n is even, n = 2k, so n(n + 1) = 2k(n + 1), which is divisible by 2.
• If n is odd, n + 1 is even, so n + 1 = 2m, and n(n + 1) = n × 2m, which is divisible by 2.
• Thus, the product is always divisible by 2.

Answer: The product of two consecutive positive integers is divisible by 2.

Solution:

Let the three consecutive integers be n, n + 1, and n + 2.

• Divisibility by 6 requires divisibility by 2 and 3.
• Divisibility by 2: Among n, n + 1, n + 2, at least one is even (e.g., if n is odd, n + 1 is even). Thus, the product is divisible by 2.
• Divisibility by 3: By Euclid’s Division Lemma, n mod 3 = 0, 1, or 2.
• If n mod 3 = 0, n is divisible by 3.
• If n mod 3 = 1, n + 2 mod 3 = 0, so n + 2 is divisible by 3.
• If n mod 3 = 2, n + 1 mod 3 = 0, so n + 1 is divisible by 3.
• Thus, at least one number is divisible by 3, so the product is divisible by 3.
• Since the product is divisible by both 2 and 3, it is divisible by 6.

Answer: The product of three consecutive positive integers is divisible by 6.

Solution:

Assume √11 is rational, i.e., √11 = p/q, where p and q are coprime integers and q ≠ 0.

• Square both sides: 11 = p²/q² → p² = 11q².
• Since p² is divisible by 11, p is divisible by 11 (let p = 11m).
• Substitute: (11m)² = 11q² → 121m² = 11q² → q² = 11m².
• Since q² is divisible by 11, it is divisible by 11.
• If p and q are both divisible by 11, they are not coprime, contradicting the assumption.
• Hence, √11 is irrational.

Answer: √11 is irrational.

Solution:

Assume 4 + √5 is rational, i.e., 4 + √5 = p/q, where p and q are coprime integers and q ≠ 0.

• Rearrange: √5 = p/q – 4 = (p – 4q)/q.
• Since p, q, and 4 are integers, (p – 4q)/q is rational.
• But √5 is irrational, so (p – 4q)/q cannot be rational.
• This contradicts the assumption, so 4 + √5 is irrational.

Answer: 4 + √5 is irrational.

Solution:

Assume 3√2 + √7 is rational, i.e., 3√2 + √7 = p/q, where p and q are coprime integers and q ≠ 0.

• Rearrange: 3√2 = p/q – √7 → √2 = (p/q – √7)/3.
• Since p/q is rational, let p/q = a. Then √2 = (a – √7)/3.
• Rearrange: √7 = a – 3√2.
• Since √2 is irrational, 3√2 is irrational, and a – 3√2 is irrational (difference of rational and irrational).
• But √7 is irrational, leading to a contradiction.
• Hence, 3√2 + √7 is irrational.

Answer: 3√2 + √7 is irrational.

Solution:

A rational number p/q has a terminating decimal if the denominator q (in lowest form) has only prime factors 2 or 5.

• Factorize 500: 500 = 2² × 5³.
• Since the denominator has only 2 and 5 as prime factors, 17/500 is terminating.

Answer: 17/500 is a terminating decimal.

Solution:

A rational number p/q has a terminating decimal if the denominator q has only prime factors 2 or 5.

• Factorize 70: 70 = 2 × 5 × 7.
• Since the denominator has 7 as a prime factor, 29/70 is non-terminating.

Answer: 29/70 is a non-terminating decimal.

Solution:

A rational number p/q has a terminating decimal if the denominator q has only prime factors 2 or 5.

• Factorize 200: 200 = 2³ × 5².
• Since the denominator has only 2 and 5 as prime factors, 13/200 is terminating.

Answer: 13/200 is a terminating decimal.