# 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.
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)

Shahrukh said:
5 years ago

If 13x * 13y = 2028 then;

It should have been 13(x*y)= 2028

And x*y= 156

But your solution is telling that

13*13(x*y)= 2028.

x*y = 2028/ 169 = 12.

So , x*y = 12.

I mean why divided by 169 why not by 13 after taking common 13? Please tell me.

It should have been 13(x*y)= 2028

And x*y= 156

But your solution is telling that

13*13(x*y)= 2028.

x*y = 2028/ 169 = 12.

So , x*y = 12.

I mean why divided by 169 why not by 13 after taking common 13? Please tell me.

(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)

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)

Ashu said:
12 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)

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.

(4)

Rbjii said:
8 years ago

Well, many of you got confused with Prime, co-Prime, why not (2,6) and all.....!!!

Lets we will Understood what's the question is ...!!!

The product of two numbers is 2028 and their H.C.F. is 13. The number of such pairs is:

step 1 : A * B = 2028 (Let A & B makes 2028 on multiply)

step 2 : (x*13) * (y*13) = 2028 (as mentioned HCF=13 for both A & B, so I took it as A=x*13, B=y*13)

******Note : HCF in the sense, only one number should be common in both A & B, hence it is 13 none other should be...!! ( are you cleared ??)*****

step 3 : x*y=2028/13*13

step 4 : x * y = 12 ( Product of X and Y makes 12, so we will separate 12 for and x & y)

*** The Sub-multiples of 12 are (1,12) (2,6) (3,4) .....still now ok...!! (Once again remember the NOTE..plzz)****

case 1 : (1,12)

A = 1 * 13

B = 12 * 13 (Product of A & B makes 2028, fine) also only 13 is common in both A & B. Hence it is satisfied

case 2 : (3,4)

A = 13 * 3

B = 13 * 4 ( Here also only 13 is common in both A & B. Hence this also satisfied )

case 3 : (2,6)

A = 13 * 2

B = 13 * 6(also 39 * 2 )

See clearly in both A and B the common numbers are 13 as well 2. It shouldn't be according to our NOTE

Hence this is not a correct pair with HCF as 13 alone. So it is said to not co-prime

JUST FOR UNDERSTANDING BIG DESCRIPTION..!! Paper Pen practice works faster ..!! ATB

Lets we will Understood what's the question is ...!!!

The product of two numbers is 2028 and their H.C.F. is 13. The number of such pairs is:

step 1 : A * B = 2028 (Let A & B makes 2028 on multiply)

step 2 : (x*13) * (y*13) = 2028 (as mentioned HCF=13 for both A & B, so I took it as A=x*13, B=y*13)

******Note : HCF in the sense, only one number should be common in both A & B, hence it is 13 none other should be...!! ( are you cleared ??)*****

step 3 : x*y=2028/13*13

step 4 : x * y = 12 ( Product of X and Y makes 12, so we will separate 12 for and x & y)

*** The Sub-multiples of 12 are (1,12) (2,6) (3,4) .....still now ok...!! (Once again remember the NOTE..plzz)****

case 1 : (1,12)

A = 1 * 13

B = 12 * 13 (Product of A & B makes 2028, fine) also only 13 is common in both A & B. Hence it is satisfied

case 2 : (3,4)

A = 13 * 3

B = 13 * 4 ( Here also only 13 is common in both A & B. Hence this also satisfied )

case 3 : (2,6)

A = 13 * 2

B = 13 * 6(also 39 * 2 )

See clearly in both A and B the common numbers are 13 as well 2. It shouldn't be according to our NOTE

Hence this is not a correct pair with HCF as 13 alone. So it is said to not co-prime

JUST FOR UNDERSTANDING BIG DESCRIPTION..!! Paper Pen practice works faster ..!! ATB

(4)

Jagrati said:
6 years ago

Product of 2 numbers = HCF * LCM.

2028= 13 * LCM.

2028/13=LCM.

Since, LCM=156.

Factors of 156 are 2*2*39.

That is the pair is 2.

2028= 13 * LCM.

2028/13=LCM.

Since, LCM=156.

Factors of 156 are 2*2*39.

That is the pair is 2.

(4)

Sheshank said:
3 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)

Julie said:
5 years ago

Thank you for giving the clear answer @Snehesh.

(1)

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