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.

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.

Can we take a pair of 13, 1?

Why we can't use the pair (4, 3) & (12, 1). We are asked to find the total number of such pairs. So (1, 12), (3, 4), (4, 3), (12, 1)?

Another way I'm telling you without co-primes.

2028/13=156, again divide 156/13 = 12.
Now factorize 12 = 1, 12, 3, 4.
There are 4 numbers so the pair is 2.

I can't understand this. Please help me to get it.

Guys, 2 and 6 are not co-primes, that means 2 is divided only by 1 and 6 is divided by both 2 and 3.

Co primes are the numbers where they should be divided only by one, in this case, they are 3 and 4.

Only 2 pairs is correct. Know the meaning of co-prime first "two numbers are said to be coprime if their H.C.F is 1 "which suits only the (1, 12) and (3, 4) pairs.

I followed this method:

13a * 13b = 2028.
13ab = 2028.
ab = 156.

Now try to find the unique pairs (Prime factorize) forming 156:
2 * 78, 3 * 52.

So there are just 2 pairs. Hence, ans is 2.

Not getting this. Can anyone explain it to me?