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:
90 comments Page 1 of 9.

Chakri said:   4 months ago
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.
(1)

Sanjay said:   6 months ago
The pairs are:
13 and 156.
26 and 78.
39 and 52.
So, there are 3 pairs.
(8)

Akshara said:   9 months ago
@Umesh.

No, it cannot be 2 and 6 because they are not coprime. 2 and 6 have 2 as a common factor.
(6)

Shivang Dwivedi said:   9 months ago
@Veerbhadra.

Because (12*13, 13) and (13, 12*13) are considered as one unique pair, irrespective of the order of writing.
(1)

Ashu said:   10 months ago
@All
why are not we considering (6,2) ..?
It is because the HCF is 13 and numbers 13a and 13b.

if a = 2 and b = 6 it means numbers are 13*2 and 13*6.
Also;
13*2 = 13*2,
13*6 = 13*2*3,
But now the HCF is 13*2 which is contradicting the initial HCF value of 13 therefore we can't consider(2,6).. .
(3)

Sheshank said:   2 years ago
@All.

Co prime means the numbers which we take, their HCF should be 1.

If we take (2, 6) their HCF is not one so we don't consider that.
(3)

Madhu said:   3 years ago
(2, 6) are not co prime numbers.

As.
2=1*2.
6=1*6, 2*3;.

So for 6. (2) is again a factor so they can't be.

Nikhil G said:   4 years ago
@All.

1*12 = 1*3*2*2.
12*1 = 1*3*2*2.
2*6 = 3*2*2.
6*2 = 3*2*2.
3*4 = 3*2*2.
4*3 = 3*2*2.

So possible pairs are only 2.
(6)

Veerabhadra said:   4 years ago
Why not answer is 4?

(1*13, 12*13) (3*13, 4*13) (12*13, 1*13) (4*13, 3*13) these are four possible pairs.
(1)

Chaitanya said:   4 years ago
Take 13x*13y( because HCF is 13)
Let 13x*13y=2028
xy=2028/13*13
xy becomes 12.

Next find out the coprimes which should be the form of x,y hence,
coprimes are (1,12) and (3,4)or (4,3),
so the real pair will be;
(13*1,13*12)and (13*3,13*4),
Hence the answer is 2.
(1)


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