# 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:

Answer: Option

Explanation:

Let the numbers 13*a* and 13*b*.

Then, 13*a* x 13*b* = 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:

91 comments Page 1 of 10.
Parmanand Singh said:
1 month ago

It should be 3 pairs because the co-prime of product 12 is (1, 12) (2, 6) (3, 4).

So we get, 3 pairs.

So we get, 3 pairs.

(3)

Chakri said:
5 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.

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:
7 months ago

The pairs are:

13 and 156.

26 and 78.

39 and 52.

So, there are 3 pairs.

13 and 156.

26 and 78.

39 and 52.

So, there are 3 pairs.

(10)

Akshara said:
10 months ago

@Umesh.

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

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:
10 months ago

@Veerbhadra.

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

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

(1)

Ashu said:
11 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).. .

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).. .

(5)

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.

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.

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.

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

(1)

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