# 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:
1
2
3
4
Explanation:

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.

Discussion:
91 comments Page 1 of 10.

Anotny mathew said:   1 decade ago
Why not 2, 6?

Yes why not 2, 6 even when they will also give 2028 on (13*2, 13*6) ?.

Why (2, 6) ?

2,6 r not co-primes.see the notes given above.

Explain this.

2 and 6 r not co-primes bcoz 6 has 2 as factor

whereas for 3 and 4 have no common factors except 1

In that case HCF would become 26.

No of pairs are (1, 12) , (3, 4). This is given in solution. But another pair we can consider here (4, 3) ? please explain this ?.

A co-prime no. is that in which no common factor other than 1.

Like in the above proble factor will be (3,4) bcoz common factor b/w them is only 1.

But if we take factor like (2,6) then common factor b/w them will be 1 and 2. So as these are not common factor hence we will not consider this.

@Sujay: you cn take (4,3) instead of (3,4)... ans will be same.