# 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:

85 comments Page 1 of 9.
Sheshank said:
1 year 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.

(1)

Madhu said:
2 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:
3 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.

(1)

Veerabhadra said:
3 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.

Chaitanya said:
3 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.

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.

Subham said:
3 years ago

The factors that is said to be co prime only when their GSF OR HCF is 1.

Umesh said:
3 years ago

The answer is incorrect. The answer could also be 2 and 6.

Randy Ranjith said:
4 years ago

Then why we take (4, 3) 4 is not co-prime?

(1)

Julie said:
4 years ago

Thank you for giving the clear answer @Snehesh.

(1)

Shahrukh said:
4 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.

(2)

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