The LCM and the HCF of two numbers are 144 and 12 , respectively. How many such pairs of numbers are possible?

Question

The LCM and the HCF of two numbers are 144 and 12 , respectively. How many such pairs of numbers are possible? (A) 0 (B) 1 (C) 2 (D) 10

Tardigrade Admin · Accepted Answer

Given LCM =144 and HCF =12 Let the two numbers be x and &gamma;. ∴ x=12 a text and &gamma;=12 b Where a and b are co-primes. We have, x × y= LCM × HCF 12 a × 12 b=144 × 12 a b=12 ⇒(a, b)=(1,12) or (3,4) There are two pairs of numbers.

Q. The LCM and the HCF of two numbers are 144 and 12 , respectively. How many such pairs of numbers are possible?

0

1

2

10

Given $L CM = 144$ and HCF $= 12$
Let the two numbers be $x$ and $γ$ .
$∴ x = 12 a and γ = 12 b$
Where $a$ and $b$ are co-primes.
We have, $x \times y =$ LCM $\times$ HCF
$12 a \times 12 b = 144 \times 12$
$ab = 12 \Rightarrow (a, b) = (1, 12)$ or $(3, 4)$
There are two pairs of numbers.

Q. The LCM and the HCF of two numbers are 144 and 12 , respectively. How many such pairs of numbers are possible?

0

1

2

10

Given LCM=144 and HCF =12 Let the two numbers be x and γ. ∴x=12a and γ=12b Where a and b are co-primes. We have, x×y= LCM × HCF 12a×12b=144×12 ab=12⇒(a,b)=(1,12) or (3,4) There are two pairs of numbers.

Given $L CM = 144$ and HCF $= 12$
Let the two numbers be $x$ and $γ$ .
$∴ x = 12 a and γ = 12 b$
Where $a$ and $b$ are co-primes.
We have, $x \times y =$ LCM $\times$ HCF
$12 a \times 12 b = 144 \times 12$
$ab = 12 \Rightarrow (a, b) = (1, 12)$ or $(3, 4)$
There are two pairs of numbers.