# 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 2 of 10.
Veena said:
1 decade ago

@vaani

In the given problem the H.C.F is 13..so if take the pair other than co-prime(i.e., the pair 2,6)....the H.C.F will become 26...but when we take the pairs 1,12;3,4....there is no common factor other than 1.. so that we have to take the pair of co-primes....

In the given problem the H.C.F is 13..so if take the pair other than co-prime(i.e., the pair 2,6)....the H.C.F will become 26...but when we take the pairs 1,12;3,4....there is no common factor other than 1.. so that we have to take the pair of co-primes....

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.

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)

Shahrukh said:
6 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)

Dhanendra thakre said:
9 years ago

Because in this problem contain highest common factor 13 if you take 2 and 6 then problem condition is not satisfied. Because 2and 6 contain 2 is become highest factor.

13*6 13*2 have HCF 2 not 13.

And 13*1 13*12 have HCF 13.

13*6 13*2 have HCF 2 not 13.

And 13*1 13*12 have HCF 13.

Mahesh said:
6 years ago

If we take 2 and 6 then we get (13*2)*(13*6)=26*78.

HCF of 26 and 78 will be 26 it's against the given question.

When we took 3 and 4 we get HCF as 13 only.

That's the reason why we shouldn't take 2 and 6.

HCF of 26 and 78 will be 26 it's against the given question.

When we took 3 and 4 we get HCF as 13 only.

That's the reason why we shouldn't take 2 and 6.

Sai ravi teja vallabhuni said:
8 years ago

Guys, 2 and 6 are not co-primes, that means 2 is divided only by 1 and 6 is divided by both 2 and 3.

Co primes are the numbers where they should be divided only by one, in this case, they are 3 and 4.

Co primes are the numbers where they should be divided only by one, in this case, they are 3 and 4.

Himank said:
9 years ago

@Naheem.

According to the formula.

Product = HCF*LCM.

Product is given 2028.

And HCF is 13.

So we get LCM = 2028/13 = 156.

Now factors of 156 is calculated which comes out to be 2*2*3*13.

According to the formula.

Product = HCF*LCM.

Product is given 2028.

And HCF is 13.

So we get LCM = 2028/13 = 156.

Now factors of 156 is calculated which comes out to be 2*2*3*13.

Snehesh said:
8 years ago

I followed this method:

13a * 13b = 2028.

13ab = 2028.

ab = 156.

Now try to find the unique pairs (Prime factorize) forming 156:

2 * 78, 3 * 52.

So there are just 2 pairs. Hence, ans is 2.

13a * 13b = 2028.

13ab = 2028.

ab = 156.

Now try to find the unique pairs (Prime factorize) forming 156:

2 * 78, 3 * 52.

So there are just 2 pairs. Hence, ans is 2.

Shubham said:
7 years ago

ab=12, now possible solutions can be 1*12, 3*4, 2*6 but we won't consider 2*6 because it will lead to HCF 13*2, as 2 is a factor of both 2 & 6. Hope you get it.

Shobs said:
8 years ago

Only 2 pairs is correct. Know the meaning of co-prime first "two numbers are said to be coprime if their H.C.F is 1 "which suits only the (1, 12) and (3, 4) pairs.

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