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

Given :

The product of two numbers is 2028 and their HCF is 13.
 

To find :

We have to find the number of such pairs.

Solution :

Let the two numbers be 'a' and 'b.

The product of the two numbers is 2028.

So, $a \times b = 2028$

HCF of a and b is 13.

So, $13 \times x = a$ and $13 \times y = b$

$13x \times 13y = 2028$

$169 \times x \times y = 2028$

$x \times y = \frac{2028}{169}$

$x \times y = 12$

x and y are co-prime.

Therefore, the possible values of (x, y) are (3, 4) and (1, 12).

Then, the possible values of (a, b) are,

$13 \times 3 = 39$ and $13 \times 4 = 52$

$13 \times 1 = 13$ and $13 \times 12 = 256$

Therefore, (39, 52) and (13, 256) are such pairs of numbers.

The number of such pairs is 2.

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Updated on: 10-Oct-2022

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