CBSE · Class 10 · Mathematics · Real NumbersFind the LCM and HCF of the following pairs of integers and verify that $\text{LCM} \times \text{HCF} = \text{Product of the two numbers}$: (i) $26$ and $91$ (ii) $510$ and $92$.

Solution for Pair (i): 26 and 91

Step 1: Prime factorization of both numbers

• Prime factorization of $26$: $$26 = 2 \times 13$$
• Prime factorization of $91$: $$91 = 7 \times 13$$

Step 2: Find HCF and LCM

• $\text{HCF}(26, 91)$ is the product of the smallest power of each common prime factor in the numbers: $$\text{HCF} = 13$$
• $\text{LCM}(26, 91)$ is the product of the greatest power of each prime factor involved in the numbers: $$\text{LCM} = 2 \times 7 \times 13 = 182$$

Step 3: Verification

• Product of the two numbers $= 26 \times 91 = 2366$
• Product of HCF and LCM $= \text{HCF} \times \text{LCM} = 13 \times 182 = 2366$
• Since $\text{Product of numbers} = \text{HCF} \times \text{LCM}$, the relation is verified.

Solution for Pair (ii): 510 and 92

Step 1: Prime factorization of both numbers

• Prime factorization of $510$: $$510 = 2 \times 3 \times 5 \times 17$$
• Prime factorization of $92$: $$92 = 2^2 \times 23 = 2 \times 2 \times 23$$

Step 2: Find HCF and LCM

• $\text{HCF}(510, 92)$ is the product of the smallest power of each common prime factor: $$\text{HCF} = 2$$
• $\text{LCM}(510, 92)$ is the product of the greatest power of each prime factor involved: $$\text{LCM} = 2^2 \times 3 \times 5 \times 17 \times 23 = 4 \times 3 \times 5 \times 17 \times 23 = 23460$$

Step 3: Verification

• Product of the two numbers $= 510 \times 92 = 46920$
• Product of HCF and LCM $= \text{HCF} \times \text{LCM} = 2 \times 23460 = 46920$
• Since $\text{Product of numbers} = \text{HCF} \times \text{LCM}$, the relation is verified.