Question

The product of two numbers is 2028 and their HCF is 13. The number of such pairs is :[A]1[B]2[C]3[D]4

Answer 1

Answer:

The product of two numbers is 2028 and their HCF is 13. The number of such pairs is :

[A]1

[B]2

[C]3

[D]4

2

Here, HCF = 13

Let the numbers be 13x and 13y, where x and y are prime to each other.

Now, 13x \times 13y = 2028

=> xy = \frac{2028}{13\times 13} = 12

The opposite pairs are : (1, 12), (3, 4), (2, 6)

But the 2 and 6 are not co-prime.

So, required no. of pairs = 2

Hence option [B] is correct answer.

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