Aptitude - Problems on H.C.F and L.C.M - Discussion

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

(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.

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.

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.

Guys you don't need to know what a coprimes for this sum. Just understand that the HCF given is 13.

Now if you take (2, 6) as another pair then the two numbers become 13*2 and 13*6.

Now the above two numbers have an HCF of 26 (because in 2 and 6 the common factor is 2) which does not satisfy the condition of the two numbers having HCF of 13 given in the question.

Hope you understand. :).

A co-prime no. is that in which no common factor other than 1.

Like in the above proble factor will be (3,4) bcoz common factor b/w them is only 1.

But if we take factor like (2,6) then common factor b/w them will be 1 and 2. So as these are not common factor hence we will not consider this.

@Sujay: you cn take (4,3) instead of (3,4)... ans will be same.

The co prime numbers those numbers which are having only one common factor that is 1
in (2&6) the factors are,

2 = 1,2.
6 = 1,2,3.

Here in 2&6 there are two factors ie, 1&2 hence they are not co primes hence it can't be used here.

Let us take 4&3,
4=1,2,4,
3=1,3,

Here only one common factor ie,1 and same for 12&1.

@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).. .

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.

I think the question have its answer.

The product of two numbers is 2028.

The Number of such pairs is 2.

Whatever it is, only 2 pairs, because we have only a and b, if we have a, b, c, d then we may have more pairs.

2 x 6
3 x 4
4 x 3
1 x 12
12 x 1
6 x 2

Correct me if I'm wrong.