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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.
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.
Kanika said:   1 decade ago
When we assume 13a AND 13b as numbers please explain.
Dinesh said:   1 decade ago
Why to take the pairs of 12?
Dhanendra thakre said:   1 decade ago
Because in this problem contain highest common factor 13 if you take 2 and 6 then problem condition is not satisfied. Because 2and 6 contain 2 is become highest factor.

13*6 13*2 have HCF 2 not 13.

And 13*1 13*12 have HCF 13.
Shefali said:   1 decade ago
If we divide 156 by 13 then it is 12. So there are 2 pairs.
SRI said:   1 decade ago
@Veena.

I too had the doubt of 2,6.... Nice clarification given. Never though to check the HCF formed by the numbers (2, 6).
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