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

Given:

Product of two numbers = 2028
H.C.F. (Highest Common Factor) = 13

Let the two numbers be:
a= 13x, b = 13y.
a= 13x, b = 13y.

Since their HCF is 13, the remaining parts xx and yy must be co-prime (i.e., gcd⁡(x,y)=1gcd(x,y)=1).

Now,
ab = 13x⋅13y = 169xy = 2028⇒xy = 2028169.
ab = 13x⋅13y = 169xy = 2028⇒xy = 1692028​.

Calculate:
2028169 = 12⇒xy = 12,
1692028​ = 12⇒xy = 12,

So, we now need to find the number of co-prime pairs (x,y)(x,y) such that xy=12xy=12 and, gcd⁡(x,y)=1gcd(x,y)=1.

Step: List all co-prime pairs with product 12

First, find all positive integer pairs (x,y)(x,y) such that xy=12xy = 12:

(1,12)(1,12)
(12,1)(12,1)
(2,6)(2,6)
(6,2)(6,2)
(3,4)(3,4)
(4,3)(4,3)

Now eliminate pairs where gcd⁡(x,y)≠1gcd(x,y)=1:
(2,6)(2,6) → gcd⁡=2gcd=2 ,wrong.
(6,2)(6,2) → gcd⁡=2gcd=2, wrong.
(1,12)(1,12) → gcd⁡=1gcd=1, correct.
(12,1)(12,1) → gcd⁡=1gcd=1, correct.
(3,4)(3,4) → gcd⁡=1gcd=1, correct.
(4,3)(4,3) → gcd⁡=1gcd=1, correct.

So, valid co-prime pairs:
(1,12),(12,1),(3,4),(4,3)(1,12),(12,1),(3,4),(4,3).

Final Step: Multiply each x and y by 13 to get the original numbers.

So, there are 4 such pairs.

Why can't we take pairs as four because (12, 1) and (4, 3) count as well right?

Anyone please explain.

It should be 3 pairs. So, the answer will be option C.

It should be 3 pairs because the co-prime of product 12 is (1, 12) (2, 6) (3, 4).

So we get, 3 pairs.

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.

The pairs are:
13 and 156.
26 and 78.
39 and 52.
So, there are 3 pairs.

@Umesh.

No, it cannot be 2 and 6 because they are not coprime. 2 and 6 have 2 as a common factor.

@Veerbhadra.

Because (12*13, 13) and (13, 12*13) are considered as one unique pair, irrespective of the order of writing.

@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).. .

@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.