We express 867 and 255 as products of prime factors: 867 = 3 × 17² and 255 = 3 × 5 × 17. The HCF is the product of the smallest power of each common prime factor involved, which is 3^1 × 17^1 = 51.

We express 867 and 255 as products of prime factors: 867 = 3 × 17² and 255 = 3 × 5 × 17. The HCF is the product of the smallest power of each common prime factor involved, which is 3^1 × 17^1 = 51.

Writing both numbers as products of their prime factors gives 96 = 2^5 × 3 and 404 = 2² × 101. The only common prime factor is 2, and its smallest power is 2² = 4, so HCF(96, 404) = 4.

Using the relation HCF(a, b) × LCM(a, b) = a × b, we compute LCM(96, 404) = (96 × 404)/(4) = 96 × 101 = 9696.

To check divisibility by 36, we find the prime factors of each given number: 12 = 2² × 3, 18 = 2 × 3², and 45 = 3² × 5. The divisor 36 factorises as 2² × 3².

The LCM is the product of the greatest power of each prime factor involved: the highest power of 2 is 2² (from 12), of 3 is 3² (from 18 and 45), and of 5 is 5^1 (from 45). Thus, LCM(12, 18, 45) contains 2² × 3² = 36 as a factor, which proves that the LCM is divisible by 36 without evaluating the complete product.

The LCM is the product of the greatest power of each prime factor involved: the highest power of 2 is 2² (from 12), of 3 is 3² (from 18 and 45), and of 5 is 5^1 (from 45). Thus, LCM(12, 18, 45) contains 2² × 3² = 36 as a factor, which proves that the LCM is divisible by 36 without evaluating the complete product.

Hcf or 867 and 255 find HCF and LCM of 96 and 404 without actually calculating the LCM determine whether the LCM of 12 18 45 is divisible by 36

Answer: 1. $\text{HCF}(867, 255) = 51$ 2. $\text{HCF}(96, 404) = 4$ , $\text{LCM}(96, 404) = 9696$ 3. Yes, $\text{LCM}(12, 18, 45)$ is divisible by $36$ .

Step-by-step solution

Step 1: Find HCF of 867 and 255 by prime factorisation

We express $867$ and $255$ as products of prime factors: $867 = 3 \times 17^2$ and $255 = 3 \times 5 \times 17$ . The HCF is the product of the smallest power of each common prime factor involved, which is $3^1 \times 17^1 = 51$ .

Step 2: Find HCF of 96 and 404

Writing both numbers as products of their prime factors gives $96 = 2^5 \times 3$ and $404 = 2^2 \times 101$ . The only common prime factor is $2$ , and its smallest power is $2^2 = 4$ , so $\text{HCF}(96, 404) = 4$ .

Step 3: Find LCM of 96 and 404 using the relationship

Using the relation $\text{HCF}(a, b) \times \text{LCM}(a, b) = a \times b$ , we compute $\text{LCM}(96, 404) = \frac{96 \times 404}{4} = 96 \times 101 = 9696$ .

Step 4: Prime factorise 12, 18, 45, and 36

To check divisibility by $36$ , we find the prime factors of each given number: $12 = 2^2 \times 3$ , $18 = 2 \times 3^2$ , and $45 = 3^2 \times 5$ . The divisor $36$ factorises as $2^2 \times 3^2$ .

Step 5: Determine divisibility of LCM(12, 18, 45) by 36

The LCM is the product of the greatest power of each prime factor involved: the highest power of $2$ is $2^2$ (from $12$ ), of $3$ is $3^2$ (from $18$ and $45$ ), and of $5$ is $5^1$ (from $45$ ). Thus, $\text{LCM}(12, 18, 45)$ contains $2^2 \times 3^2 = 36$ as a factor, which proves that the LCM is divisible by $36$ without evaluating the complete product.

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