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
Answer: Option
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:
94 comments Page 8 of 10.

Julie said:   7 years ago
Thank you for giving the clear answer @Snehesh.
(1)

Raj said:   1 decade ago
2,6 r not co-primes.see the notes given above.

Disha said:   8 years ago
Not getting this. Can anyone explain it to me?

Randy Ranjith said:   6 years ago
Then why we take (4, 3) 4 is not co-prime?
(1)

Sriram said:   1 decade ago
We have to consider the co primes with 12.

Gokul said:   7 years ago
xy=2028/13=156.

Then how to solve it?

Donald said:   1 decade ago
It's hard to understand not satisfied.

Sai said:   10 years ago
13a*13b = 2028.
169ab = 2028.
ab = 12.

Subbu said:   10 years ago
I am not getting the concept of ab=12.

Anik said:   8 years ago
Why not 2, 6? Please explain in clear.


Post your comments here:

Your comments will be displayed after verification.