Question

hcf of two numbers is 21 and their product is 2646 how many pairs of such numbers are possible

Answer 1

let the number be 13a and 13b.
then,13a ×13b=2028
then,ab=12
the co-primes with product 12 are (1,12) (3,4)
So the required numbers are (13×1),(13×12) , (13×3) and (13×4)
The hcf is 13.
there are 2 such pairs.

Answer 2

Answer:

2

Step-by-step explanation:

Since 21 is a factor of both numbers let the numbers be 21a and 21b.

product of 2 numbers = HCF x LCM

21a x 21b = 2646

=> a x b = 2646/21 x 21

=> a x b = 6

Co primes with product 6 are (1,6) and (3,2).

therefore the numbers are (21,126) and (63,42)

Hence, 2 pairs.