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 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?
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.
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.
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