Prime factorisation of 867 = 3 × 17 × 17 = 3 × 17² Prime factorisation of 255 = 3 × 5 × 17 Common prime factors = 3 and 17 ∴ HCF = 3 × 17 = 51 Now, LCM = (867 × 255) / 51 = 4335 Verification: 51 × 4335 = 867 × 255 Let the prime factorisation of a be as follows: a = p₁ × p₂ × ... × pₙ where p₁, p₂, ..., pₙ are prime numbers, not necessarily distinct. Therefore, a² = (p₁ × p₂ × ... × pₙ)(p₁ × p₂ × ... × pₙ) = p₁² × p₂² × ... × pₙ² Now, we are given that p divides a². So, by the Fundamental Theorem of Arithmetic, p must be one of the prime factors of a². But the prime factors of a² are exactly p₁, p₂, ..., pₙ (using the uniqueness of prime factorisation). Hence, p must be one of p₁, p₂, ..., pₙ. Since a = p₁ × p₂ × ... × pₙ, it follows that p divides a. ∴ If p divides a², then p divides a.Let us assume, to the contrary, that √7 is rational. Then, we can find coprime integers a and b (b ≠ 0) such that √7 = a/b. Rearranging, 7 = a²/b². Multiplying both sides by b², we get 7b² = a². This shows that a² is a multiple of 7. By the property of prime factorisation, if a prime number divides a square, it must divide the number itself. Hence, 7 must divide a. Let a = 7k for some integer k. Substituting in 7b² = a², we get 7b² = (7k)² = 49k². Dividing both sides by 7, we get b² = 7k². This shows that b² is a multiple of 7. By the same property of prime factorisation, if a prime number divides a square, it must divide the number itself. Hence, 7 must divide b. But this contradicts our assumption that a and b are coprime (they have no common factors other than 1). Therefore, our assumption is wrong. ∴ sqrt(7) is irrational.For a number to end with the digit 0, it must be divisible by 10. This means it must be divisible by both 2 and 5. The prime factorisation of 7ⁿ = (7)ⁿ = 7 × 7 × ... × 7 (n times). The only prime factor in the factorisation of 7ⁿ is 7. It does not contain the prime factors 2 and 5. By the Fundamental Theorem of Arithmetic, the prime factorisation of a number is unique. Therefore, 7ⁿ cannot have 2 and 5 as factors. Hence, 7ⁿ cannot end with the digit 0 for any natural number n.8 × 7 × 6 × 5 × 4 × 3 × 2 + (1+5) 1 + 5 = 6 Expression = (8 × 7 × 6 × 5 × 4 × 3 × 2) + 6 Taking 6 common: = 6 × (8 × 7 × 5 × 4 × 3 × 2 + 1) Since the number has a factor 6 (other than 1 and itself), it is a composite number. (7 × 5 × 4 × 3) + (1295/37) 1295 ÷ 37 = 35 Expression = (7 × 5 × 4 × 3) + 35 Since 35 = 7 × 5, we can write: = (7 × 5 × 4 × 3) + (7 × 5) Taking (7 × 5) common: = 35 × (4 × 3 + 1) = 35 × (12 + 1) = 35 × 13 Since the number has factors 35 and 13 (other than 1 and itself), it is a composite number.Prime factorisation: 306 = 2 × 3² × 17 657 = 3² × 73 HCF = 3² = 9 LCM = 2 × 3² × 17 × 73 = 22338 Verification: HCF × LCM = 9 × 22338 = 201042 306 × 657 = 201042 Hence verified.To find when they will leave together again, we need to find the LCM of their intervals. Intervals: 15, 20, 30 minutes. Prime factorisation: 15 = 3 × 5 20 = 2² × 5 30 = 2 × 3 × 5 LCM(15, 20, 30) = 2² × 3 × 5 = 4 × 3 × 5 = 60 minutes. So, the buses will leave together again after 60 minutes (1 hour). If they left at 8:00 AM, they will next leave together at 8:00 AM + 1 hour = 9:00 AM. Answer: 9:00 AM.Assume, to the contrary, that √3 + √5 = r, where r is a rational number. Squaring both sides: (√3 + √5)² = r² 3 + 5 + 2√15 = r² 8 + 2√15 = r² Rearranging: 2√15 = r² - 8 √15 = (r² - 8) / 2 Since r is rational, r² - 8 is rational, hence (r² - 8) / 2 is rational. This implies √15 is rational, which is impossible because 15 is not a perfect square. Let us assume, to the contrary, that √3 + √5 is rational. Then, we can find coprime integers a and b (b ≠ 0) such that √3 + √5 = a/b. Squaring both sides: (√3 + √5)² = (a/b)² 3 + 5 + 2√15 = a²/b² 8 + 2√15 = a²/b² 2√15 = a²/b² - 8 √15 = (a²/b² - 8) / 2 √15 = (a² - 8b²) / 2b² Since a and b are integers, a² - 8b² and 2b² are also integers. Thus, (a² - 8b²) / 2b² is a rational number. This implies that √15 is rational. But we know that √15 is irrational. This contradicts the fact that √15 is irrational. Hence, our assumption is wrong. ∴ √3 + √5 is irrational. Assume, to the contrary, that √3 + √5 = r, where r is a rational number. √3 = r - √5 Squaring both sides: 3 = (r - √5)² 3 = r² - 2r√5 + 5 2r√5 = r² + 2 √5 = (r² + 2) / 2r Since r is rational, r² + 2 is rational, hence (r² + 2) / 2r is rational. This implies √5 is rational, which is impossible because 5 is not a perfect square. Hence, our assumption is wrong. ∴ √3 + √5 is irrational.To find the greatest number that will divide the given numbers leaving the specified remainders, we need to find the HCF of the differences between the numbers and their remainders. Differences: 4786 - 6 = 4780, 5962 - 10 = 5952, 7148 - 14 = 7134. Prime factorisation: 4780 = 2² × 5 × 239 5952 = 2⁵ × 3² × 11 7134 = 2 × 3² × 7 × 173 HCF is the product of the lowest power of each common prime factor in the numbers. HCF(4780, 5952, 7134) = 2 So, the greatest number that will divide the given numbers leaving the specified remainders is 2.To find the smallest number which when decreased by 11 is divisible by 28, 35, and 42, we need to find the LCM of 28, 35, and 42. LCM is the product of the highest power of each prime factor involved in the numbers. Prime factorisation: 28 = 2² × 7 35 = 5 × 7 42 = 2 × 3 × 7 LCM(28, 35, 42) = 2² × 3 × 5 × 7 = 4 × 3 × 5 × 7 = 420. So, the smallest number which when decreased by 11 is divisible by 28, 35, and 42 is 420 + 11 = 431.To find the length of the plank which can be used to measure exactly the lengths 4 m 50 cm, 9 m 90 cm, and 16 m 20 cm in the least time, we need to find the HCF of 4 m 50 cm, 9 m 90 cm, and 16 m 20 cm. HCF is the product of the lowest power of each common prime factor in the numbers. Prime factorisation: 4 m 50 cm = 450 cm 9 m 90 cm = 990 cm 16 m 20 cm = 1620 cm HCF(450, 990, 1620) = 30 cm. So, the length of the plank which can be used to measure exactly the lengths 4 m 50 cm, 9 m 90 cm, and 16 m 20 cm in the least time is 30 cm.Assume, to the contrary, that 6√7/5 is a rational number. Then, we can find coprime integers a and b (b ≠ 0) such that 6√7/5 = a/b. √7 = 5a/6b. Since a and b are integers, 5a and 6b are also integers. Hence, 5a/6b is a rational number. This contradicts the fact that √7 is irrational. Hence, our assumption is wrong. ∴ 6√7/5 is irrational.x=(2+√3),so 1/x=1/(2+√3),Rationalise by multiplying by (2-√3)/(2-√3) 1/x=(2-√3)/(2-√3)(2+√3)/(2+√3) 1/x=(2-√3)/4-3 1/x=(2-√3)/1 1/x=(2-√3) x-1/x=2+√3-(2-√3) x-1/x=2+√3-2+√3 x-1/x=2√3 x-1/x is irrational because √3 is irrational. Prime factorization: 36 = 2² × 3² 48 = 2⁴ × 3 60 = 2² × 3 × 5 HCF is the product of the lowest power of each common prime factor in the numbers. HCF of 36, 48, and 60 is 2² × 3 = 12. LCM is the product of the highest power of each prime factor involved in the numbers. LCM of 36, 48, and 60 is 2⁴ × 3² × 5 = 144. Product of the three numbers = 36 × 48 × 60 = 103680. Product of their HCF and LCM = 12 × 144 = 1728. Since 103680 is not equal to 1728, the product of the three numbers is not equal to the product of their HCF and LCM.Prime factorization of 10800: 10800 = 108 × 100 108 = 2² × 3³ 100 = 2² × 5² So, 10800 = 2² × 3³ × 2² × 5² = 2⁴ × 3³ × 5² Now, √10800 = √(2⁴ × 3³ × 5²) √10800 = √(2⁴ × 3² × 3 × 5²) √10800 = (2² × 3 × 5) × √3 √10800 = 60√3 Since √3 is an irrational number and the product of a non-zero rational number (60) and an irrational number (√3) is irrational, Therefore, √10800 is an irrational number.
