Problems on HCF and LCM

Q1: The product of two numbers is 2028 and their H.C.F. is 13. The number of such pairs is:

A 3

B 4

C 2

D 1

ANS:B - 2

Let the numbers 13a and 13b. Then, 13a x 13b = 2028  ab = 12. Now, the co-primes with product 12 are (1, 12) and (3, 4). [Note: Two integers a and b are said to be coprime or relatively prime if they have no common positive factor other than 1 or, equivalently, if their greatest common divisor is 1 ] So, the required numbers are (13 x 1, 13 x 12) and (13 x 3, 13 x 4). Clearly, there are 2 such pairs.



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