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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 8 of 10.
Priyanka said:   1 decade ago
Yes why not 2, 6 even when they will also give 2028 on (13*2, 13*6) ?.
Shefali said:   1 decade ago
If we divide 156 by 13 then it is 12. So there are 2 pairs.
Sai said:   10 years ago
13a*13b = 2028.
169ab = 2028.
ab = 12.
Priya said:   10 years ago
I am not getting the concept of ab=12?

How it came an some one please explain?
Himank said:   10 years ago
@Naheem.

According to the formula.

Product = HCF*LCM.

Product is given 2028.

And HCF is 13.

So we get LCM = 2028/13 = 156.

Now factors of 156 is calculated which comes out to be 2*2*3*13.
Deepak said:   10 years ago
4 is definitely not prime number, then why you have taken as co prime?

Only one pair can be done i.e. 13, 1.
Naheem sheikh said:   10 years ago
x*y = 2028.
x*13 = 2028.
x = 156.

Factor = 13*12.

Now, please explain? Anyone in details?
Deepika said:   10 years ago
Can anyone explain in a simple way?
Yash Srivastava said:   10 years ago
HCF of 26 and 78 is 2 not 13.
Donald said:   1 decade ago
It's hard to understand not satisfied.
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