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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)
Problems on H.C.F and L.C.M
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 6 of 10.
Naheem sheikh said:   1 decade ago
x*y = 2028.
x*13 = 2028.
x = 156.

Factor = 13*12.

Now, please explain? Anyone in details?
Samaria vincent said:   1 decade ago
First of all why not 5, 7 because even these numbers are co-prime numbers. Isn't it?
Priya said:   1 decade ago
I am not getting the concept of ab=12?

How it came an some one please explain?
Chinnu said:   1 decade ago
(3, 4)4 = 1, 2, 4 why it is coprime number 2 number is also in the factor of 4.
Pradeep said:   9 years ago
Why they take the two pair like (1, 12), (3, 4)?

Why not (12, 1), (4, 3)?
Nagendra said:   8 years ago
In the question, they doesn't mention about the co-prime numbers, right?
Subham said:   5 years ago
The factors that is said to be co prime only when their GSF OR HCF is 1.
Sanjay said:   2 years ago
The pairs are:
13 and 156.
26 and 78.
39 and 52.
So, there are 3 pairs.
(22)
Priyanka said:   1 decade ago
Yes why not 2, 6 even when they will also give 2028 on (13*2, 13*6) ?.
Pallavi said:   1 decade ago
Got the point veena, sorry to post the earlier comment. Thank !
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