# 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 8 of 10.
Sathish kumar said:
1 decade ago

(a*b)product of number = l.c.m * h.c.f

2028 = x * 13

x = 156.

Which means

l.c.m = 156

h.c.f = 13

Now

156 = 2 x 6 x 13

Factorize the HCF

13 = 1 * 13

The HCF is the common factor for both numbers whereas the LCM contains all the unique factors of both numbers.

So, 13 is a common factor for both numbers.

The factors remaining are 2 and 6.

So, 2 is a factor of one number and 6 is the factor of the second number.

The first number = 13 x 2 = 26

The second number = 13 x 6 = 78

Hope it is clear now.

Now

Factorize the 26 and 78

26 = 2 * 13

78 = 6 * 13

Here 13 is the common value

So h.c.f. is 13 as given in question.

And

156 is the l.c.m.

And finally the number of pair is 1(13 only)

Then how answer as two(2).

Tell me clearly, how to solve, its my method is correct.

2028 = x * 13

x = 156.

Which means

l.c.m = 156

h.c.f = 13

Now

156 = 2 x 6 x 13

Factorize the HCF

13 = 1 * 13

The HCF is the common factor for both numbers whereas the LCM contains all the unique factors of both numbers.

So, 13 is a common factor for both numbers.

The factors remaining are 2 and 6.

So, 2 is a factor of one number and 6 is the factor of the second number.

The first number = 13 x 2 = 26

The second number = 13 x 6 = 78

Hope it is clear now.

Now

Factorize the 26 and 78

26 = 2 * 13

78 = 6 * 13

Here 13 is the common value

So h.c.f. is 13 as given in question.

And

156 is the l.c.m.

And finally the number of pair is 1(13 only)

Then how answer as two(2).

Tell me clearly, how to solve, its my method is correct.

Anup said:
1 decade ago

Is there a simpler method to solve this?

This was my solution but I don't know why is it wrong.

13 is the H.C.F.

So the numbers should have 13 in them. Also as a*b = 12.

We can have these combinations.

(13,13*2*2*4) (13*2,13*2*3) (13*3,13*2*2)

Please correct me if my logic is not appropriate.

This was my solution but I don't know why is it wrong.

13 is the H.C.F.

So the numbers should have 13 in them. Also as a*b = 12.

We can have these combinations.

(13,13*2*2*4) (13*2,13*2*3) (13*3,13*2*2)

Please correct me if my logic is not appropriate.

S ADITYA GAUTAM said:
1 decade ago

We take either (1,12) or (4,3) as coprimes and not (2,6) becouse of two reasons:

1)we want to get the product 12 as ab=12 only.

2)we choose coprimes (1,12) or (3,4) not non-coprimes(2,6) because if we choose the non-coprimes(2,6) then we get two nos as (26,78) whose H.C.F is 26 not 13 which against the question.

On the other hand if coprime numbers are taken like(1,12) then we get the two nos. as (13,156) and for coprime pairs(3,4) we get the numbers (39,52).In both of the cases u may observe that the H.C.F is 13 only.

1)we want to get the product 12 as ab=12 only.

2)we choose coprimes (1,12) or (3,4) not non-coprimes(2,6) because if we choose the non-coprimes(2,6) then we get two nos as (26,78) whose H.C.F is 26 not 13 which against the question.

On the other hand if coprime numbers are taken like(1,12) then we get the two nos. as (13,156) and for coprime pairs(3,4) we get the numbers (39,52).In both of the cases u may observe that the H.C.F is 13 only.

Samaria vincent said:
1 decade ago

First of all why not 5, 7 because even these numbers are co-prime numbers. Isn't it?

RAHUL KUMAR SARASWAT said:
1 decade ago

Sir we have a formula that the product of two numbers= LCM * HCM.

Why this formula is not used in this question ?

Why this formula is not used in this question ?

Pallavi said:
1 decade ago

Got the point veena, sorry to post the earlier comment. Thank !

Pallavi said:
1 decade ago

Why not (13*6, 13*2) be another pair?

Aajush said:
1 decade ago

Its not compulsory to get from prime or co-prime.

The thing is to get it not exceeding 2 pair.

The thing is to get it not exceeding 2 pair.

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....

Utsavchavda said:
1 decade ago

Meaning of co-prime ?

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