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

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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