# Aptitude - Problems on H.C.F and L.C.M - Discussion

Discussion Forum : Problems on H.C.F and L.C.M - General Questions (Q.No. 9)

9.

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

Answer: Option

Explanation:

Let the numbers 13*a* and 13*b*.

Then, 13*a* x 13*b* = 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.

Discussion:

91 comments Page 10 of 10.
Parmanand Singh said:
3 months ago

It should be 3 pairs because the co-prime of product 12 is (1, 12) (2, 6) (3, 4).

So we get, 3 pairs.

So we get, 3 pairs.

(6)

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