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 5 of 10.

Yash Srivastava said:   1 decade ago
HCF of 26 and 78 is 2 not 13.

Deepika said:   1 decade ago
Can anyone explain in a simple way?

Naheem sheikh said:   1 decade ago
x*y = 2028.
x*13 = 2028.
x = 156.

Factor = 13*12.

Now, please explain? Anyone in details?

Deepak said:   1 decade 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.

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.

Priya said:   10 years ago
I am not getting the concept of ab=12?

How it came an some one please explain?

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.

Xing said:   10 years ago
Longer method, divide both side by 13.

=> 13a * 13b/13 = 2028/13.

=> 13a * b = 156.

=> ab = 156/13.

=> ab = 12.
(1)

Vinay ambidi said:   10 years ago
No someone asked why 5, 7 but if we multiply those numbers we will get 5*7 = 35.

That's why we have to take 1, 12 or 3, 4.


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