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 4 of 10.

Pradeep said:   10 years ago
Why they take the two pair like (1, 12), (3, 4)?

Why not (12, 1), (4, 3)?

Sami said:   10 years ago
How to choose the pairs?

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.

Umesh said:   6 years ago
The answer is incorrect. The answer could also be 2 and 6.

Anik said:   9 years ago
Why not 2, 6? Please explain in clear.

Disha said:   8 years ago
Not getting this. Can anyone explain it to me?

Balachandra Padiyar said:   8 years ago
The co prime numbers those numbers which are having only one common factor that is 1
in (2&6) the factors are,

2 = 1,2.
6 = 1,2,3.

Here in 2&6 there are two factors ie, 1&2 hence they are not co primes hence it can't be used here.

Let us take 4&3,
4=1,2,4,
3=1,3,

Here only one common factor ie,1 and same for 12&1.

Shivangi said:   8 years ago
But Question is how many no of the pair. Not co factors. Please clear my doubt is no of pair means co-factor?

Kishan said:   8 years ago
The product of two numbers is 2028 and their HCF is 13 If the numbers of two digits find the numbers.

Can anyone answer me?

Nagendra said:   8 years ago
In the question, they doesn't mention about the co-prime numbers, right?


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