# 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:
1
2
3
4
Explanation:

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.

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)

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

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.

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.

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 ?

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

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

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

The thing is to get it not exceeding 2 pair.