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

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.

@Veerbhadra.

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

Then why we take (4, 3) 4 is not co-prime?

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.

Thank you for giving the clear answer @Snehesh.

LCM=2028/13=156 which means 13x12 or 12*13 so there are 2 ways.

Longer method, divide both side by 13.

=> 13a * 13b/13 = 2028/13.

=> 13a * b = 156.

=> ab = 156/13.

=> ab = 12.

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.

(2, 6) are not co prime numbers.

As.
2=1*2.
6=1*6, 2*3;.

So for 6. (2) is again a factor so they can't be.

Why not answer is 4?

(1*13, 12*13) (3*13, 4*13) (12*13, 1*13) (4*13, 3*13) these are four possible pairs.