The product of two numbers is 5292 and their H.C.F. is 21. The number of such pairs is:

Correct option is C

​Given:

Product of two numbers = 5292

HCF = 21

Concept Used:

If HCF = h, then the numbers can be written as h × a and h × b, where hcf⁡(a,b) = 1

So,

(h × a)(h × b) = h^2​ × ab

Formula Used:

Number of coprime pairs (a,b) such that ab = $$\frac{\text{Product}}{(\text{HCF})^2}$$​​

If ab = N, then number of such coprime pairs = $$\frac{1}{2} \times 2^n$$​, where n is the number of distinct prime factors of N

Solution:

Now,

​$$(h \times a)(h \times b) = 5292$$​​

​$$h^2 \times ab = 5292$$​​

​$$(21)^2 \times ab = 5292$$​​

ab = $$\frac{5292}{21^2}$$​ = 12

Factors of 12: 1,2,3,4,6,12

Coprime pairs with product 12: (1,12),(3,4) - 2 pairs