Question

2. For each of the following pairs of numbers, show that the product of their HCF and LCM equals their product: (i) 14,21 (ii) 27,90​

Answer 1

HCF of 14,21 can be calculated as follows

14=2×7

21=3×7

As 7 is common in both HCF=7 now

LCM=2×3×7 product of all factors appear in the number but only once

Therefore,LCM=42

Now HCF ×LCM=7×42=294…(1)

Now 14×21=294…(2)

From (1) and(2)

LHS=RHS

Hence proved

2) 27,90

27=3×3×3

90=3×3×2×5

Now two three are common

Therefore HCF= 3×3=9

LCM=common factor × uncommon factor=[3×3]×[3×2×5]=270

Now HCF×LCM=9×270=2430…(1)

27×90=2430…(2)

From (1) and (2)

LHS=RHS

hence proved

Answer 2

Question:

• For each of the following pairs of numbers, show that the product of their HCF and LCM equals their product:

(i) 14, 21

By using prime factorisation method.

• Factors of 14 = 2 × 7
• Factors of 21 = 3 × 7

$\dashrightarrow$HCF (14, 21) = 7

$\dashrightarrow$LCM (14, 21) = 2 × 3 × 7 = 42

HCF × LCM = Product of numbers

↠ 7 × 42 = 14 × 21

↠ 294 = 294

$\huge\bold{Hence, Verified}$

(ii) 27, 90

By using prime factorisation method.

• Factors of 27 = 3³
• Factors of 90 = 2 × 3² × 5

$\dashrightarrow$HCF (27, 90) = 3² = 9

$\dashrightarrow$LCM (27, 90) = 2 × 3³ × 5 = 270

HCF × LCM = Product of numbers

↠ 9 × 270 = 27 × 90

↠ 2430 = 2430

$\huge\bold{Hence, Verified}$